Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5 Questions and Answers.

11th Maths Part 2 Methods of Induction and Binomial Theorem Exercise 4.5 Questions And Answers Maharashtra Board

Question 1.
Show that C₀ + C₁ + C₂ + ….. + C₈ = 256
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + ….. + C₈ = 2⁸
∴ C₀ + C₁ + C₂ + ….. + C₈ = 256

Question 2.
Show that C₀ + C₁ + C₂ + …… + C₉ = 512
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 9, we get
C₀ + C₁ + C₂ + ….. + C₉ = 2⁹
∴ C₀ + C₁ + C₂ + …… + C₉ = 512

Question 3.
Show that C₁ + C₂ + C₃ + ….. + C₇ = 127
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 7, we get
C₀ + C₁ + C₂ + ….. + C₇ = 2⁷
∴ C₀ + C₁ + C₂ +….. + C₇ = 128
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₇ = 128
∴ C₁ + C₂ + ….. + C₇ = 128 – 1 = 127

Question 4.
Show that C₁ + C₂ + C₃ + ….. + C₆ = 63
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 6, we get
C₀ + C₁ + C₂ + ….. + C₆ = 2⁶
∴ C₀ + C₁ + C₂ + …… + C₆ = 64
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₆ = 64
∴ C₁ + C₂ + ….. + C₆ = 64 – 1 = 63

Question 5.
Show that C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128
Solution:
Since C₀ + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + C₃ + …… + C₈ = 2⁸
But, sum of even coefficients = sum of odd coefficients
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇
Let C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = k
Now, C₀ + C₁ + C₂ + C₃ + C₄ + C₅ + C₆ + C₇ + C₈ = 256
∴ (C₀ + C₂ + C₄ + C₆ + C₈) + (C₁ + C₃ + C₅ + C₇) = 256
∴ k + k = 256
∴ 2k = 256
∴ k = 128
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128

Question 6.
Show that C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
But, C₀ = 1
∴ 1 + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
∴ C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1

Question 7.
Show that C₀ + 2C₁ + 3C₂ + 4C₃ + ….. + (n + 1)C_n = (n + 2) 2^n-1
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 4.1 Class 11 Maths 2
• Exercise 4.2 Class 11 Maths 2
• Exercise 4.3 Class 11 Maths 2
• Exercise 4.4 Class 11 Maths 2
• Exercise 4.5 Class 11 Maths 2
• Miscellaneous Exercise 4 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4 Questions and Answers.

11th Maths Part 2 Methods of Induction and Binomial Theorem Exercise 4.4 Questions And Answers Maharashtra Board

Question 1.
State, by writing the first four terms, the expansion of the following, where |x| < 1.
(i) (1 + x)^-4
Solution:

(ii) (1 – x)^1/3
Solution:

(iii) (1 – x²)^-3
Solution:

(iv) (1 + x)^-1/5
Solution:

(v) (1 + x²)^-1
Solution:

Question 2.
State by writing first four terms, the expansion of the following, where |b| < |a|.
(i) (a – b)^-3
Solution:
(a – b)^-3 = \(\left[a\left(1-\frac{b}{a}\right)\right]^{-3}\)

(ii) (a + b)^-4
Solution:

(iii) (a + b)^1/4
Solution:

(iv) (a – b)^-1/4
Solution:
(a – b)^-1/4 = \(\left[a\left(1-\frac{b}{a}\right)\right]^{\frac{-1}{4}}\)

(v) (a + b)^-1/3
Solution:

Question 3.
Simplify the first three terms in the expansion of the following:
(i) (1 + 2x)^-4
Solution:

(ii) (1 + 3x)^-1/2
Solution:

(iii) (2 – 3x)^1/3
Solution:

(iv) (5 + 4x)^-1/2
Solution:

(v) (5 – 3x)^-1/3
Solution:

Question 4.
Use the binomial theorem to evaluate the following upto four places of decimals.
(i) √99
Solution:

= 10 [1 – 0.005 – 0.0000125 – ……]
= 10(0.9949875)
= 9.94987 5
= 9.9499

(ii) \(\sqrt[3]{126}\)
Solution:

(iii) \(\sqrt[4]{16.08}\)
Solution:

(iv) (1.02)^-5
Solution:

(v) (0.98)^-3
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 4.1 Class 11 Maths 2
• Exercise 4.2 Class 11 Maths 2
• Exercise 4.3 Class 11 Maths 2
• Exercise 4.4 Class 11 Maths 2
• Exercise 4.5 Class 11 Maths 2
• Miscellaneous Exercise 4 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3 Questions and Answers.

11th Maths Part 2 Methods of Induction and Binomial Theorem Exercise 4.3 Questions And Answers Maharashtra Board

Question 1.
In the following expansions, find the indicated term.
(i) \(\left(2 x^{2}+\frac{3}{2 x}\right)^{8}\), 3rd term
Solution:

(ii) \(\left(x^{2}-\frac{4}{x^{3}}\right)^{11}\), 5th term
Solution:

(iii) \(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{9}\), 7th term
Solution:

(iv) In \(\left(\frac{1}{3}+a^{2}\right)^{12}\), 9th term
Solution:

(v) In \(\left(3 a+\frac{4}{a}\right)^{13}\), 10th term
Solution:

Question 2.
In the following expansions, find the indicated coefficients.
(i) x³ in \(\left(x^{2}+\frac{3 \sqrt{2}}{x}\right)^{9}\)
Solution:

(ii) x⁸ in \(\left(2 x^{5}-\frac{5}{x^{3}}\right)^{8}\)
Solution:

(iii) x⁹ in \(\left(\frac{1}{x}+x^{2}\right)^{18}\)
Solution:

(iv) x^-3 in \(\left(x-\frac{1}{2 x}\right)^{5}\)
Solution:

(v) x^-20 in \(\left(x^{3}-\frac{1}{2 x^{2}}\right)^{15}\)
Solution:

Question 3.
Find the constant term (term independent of x) in the expansion of
(i) \(\left(2 x+\frac{1}{3 x^{2}}\right)^{9}\)
Solution:

(ii) \(\left(x-\frac{2}{x^{2}}\right)^{15}\)
Solution:

(iii) \(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{10}\)
Solution:

(iv) \(\left(x^{2}-\frac{1}{x}\right)^{9}\)
Solution:

(v) \(\left(2 x^{2}-\frac{5}{x}\right)^{9}\)
Solution:

Question 4.
Find the middle terms in the expansion of
(i) \(\left(\frac{x}{y}+\frac{y}{x}\right)^{12}\)
Solution:

(ii) \(\left(x^{2}+\frac{1}{x}\right)^{7}\)
Solution:

(iii) \(\left(x^{2}-\frac{2}{x}\right)^{8}\)
Solution:

(iv) \(\left(\frac{x}{a}-\frac{a}{x}\right)^{10}\)
Solution:

(v) \(\left(x^{4}-\frac{1}{x^{3}}\right)^{11}\)
Solution:

Question 5.
In the expansion of (k + x)⁸, the coefficient of x⁵ is 10 times the coefficient of x⁶. Find the value of k.
Solution:

Question 6.
Find the term containing x⁶ in the expansion of (2 – x) (3x + 1)⁹.
Solution:

Question 7.
The coefficient of x² in the expansion of (1 + 2x)^m is 112. Find m.
Solution:
The coefficient of x² in (1 + 2x)^m = ^mC₂ (2²)
Given that the coefficient of x² = 112
∴ ^mC₂ (4) = 112
∴ ^mC₂ = 28
∴ \(\frac{\mathrm{m} !}{2 !(\mathrm{m}-2) !}=28\)
∴ \(\frac{m(m-1)(m-2) !}{2 \times(m-2) !}=28\)
∴ m(m – 1) = 56
∴ m² – m – 56 = 0
∴ (m – 8) (m + 7) = 0
As m cannot be negative.
∴ m = 8

Class 11 Maharashtra State Board Maths Solution 

• Exercise 4.1 Class 11 Maths 2
• Exercise 4.2 Class 11 Maths 2
• Exercise 4.3 Class 11 Maths 2
• Exercise 4.4 Class 11 Maths 2
• Exercise 4.5 Class 11 Maths 2
• Miscellaneous Exercise 4 Class 11 Maths 2

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2 Questions and Answers.

11th Maths Part 2 Methods of Induction and Binomial Theorem Exercise 4.2 Questions And Answers Maharashtra Board

Question 1.
Expand:
(i) (√3 + √2)⁴
Solution:
Here, a = √3, b = √2 and n = 4.
Using binomial theorem,

∴ (√3 + √2)⁴ = 1(9) (1) + 4(3√3) (√2) + 6(3)(2) + 4(√3) (2√2) + 1(1)(4)
= 9 + 12√6 + 36 + 8√6 + 4
= 49 + 20√6

(ii) (√5 – √2)⁵
Solution:
Here, a = √5, b = √2 and n = 5.
Using binomial theorem,

Question 2.
Expand:
(i) (2x² + 3)⁴
Solution:
Here, a = 2x², b = 3 and n = 4.
Using binomial theorem,

(ii) \(\left(2 x-\frac{1}{x}\right)^{6}\)
Solution:
Here, a = 2x, b = \(\frac{1}{x}\) and n = 6.
Using binomial theorem,

Question 3.
Find the value of
(i) (√3 + 1)⁴ – (√3 – 1)⁴
Solution:

(ii) (2 + √5)⁵ + (2 – √5)⁵
Solution:


Adding (i) and (ii), we get
∴ (2 + √5 )⁵ + (2 – √5)⁵ = (32 + 80√5 + 400 + 200√5 + 250 + 25√5) + (32 – 80√5 + 400 – 200√5+ 250 – 25√5 )
= 64 + 800 + 500
= 1364

Question 4.
Prove that:
(i) (√3 + √2)⁶ + (√3 – √2)⁶ = 970
Solution:

(ii) (√5 + 1)⁵ – (√5 – 1)⁵ = 352
Solution:

Question 5.
Using binomial theorem, find the value of
(i) (102)⁴
Solution:

(ii) (1.1)⁵
Solution:

Question 6.
Using binomial theorem, find the value of
(i) (9.9)³
Solution:

(ii) (0.9)⁴
Solution:

Question 7.
Without expanding, find the value of
(i) (x + 1)⁴ – 4(x + 1)³ (x – 1) + 6(x + 1)² (x – 1)² – 4(x + 1) (x – 1)³ + (x – 1)⁴
Solution:

(ii) (2x – 1)⁴ + 4(2x – 1)³ (3 – 2x) + 6(2x – 1)² (3 – 2x)² + 4(2x – 1)¹ (3 – 2x)³ + (3 – 2x)⁴
Solution:

Question 8.
Find the value of (1.02)⁶, correct upto four places of decimals.
Solution:

Question 9.
Find the value of (1.01)⁵, correct upto three places of decimals.
Solution:

Question 10.
Find the value of (0.9)⁶, correct upto four places of decimals.
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 4.1 Class 11 Maths 2
• Exercise 4.2 Class 11 Maths 2
• Exercise 4.3 Class 11 Maths 2
• Exercise 4.4 Class 11 Maths 2
• Exercise 4.5 Class 11 Maths 2
• Miscellaneous Exercise 4 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.1 Questions and Answers.

11th Maths Part 2 Methods of Induction and Binomial Theorem Exercise 4.1 Questions And Answers Maharashtra Board

Prove by the method of induction, for all n ∈ N.

Question 1.
2 + 4 + 6 + …… + 2n = n(n + 1)
Solution:
Let P(n) = 2 + 4 + 6 + …… + 2n = n(n + 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(1 + 1) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2 + 4 + 6 + ….. + 2k = k(k + 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2 + 4 + 6 + …… + 2(k + 1) = (k + 1) (k + 2)
L.H.S. = 2 + 4 + 6 + …+ 2(k + 1)
= 2 + 4 + 6+ ….. + 2k + 2(k + 1)
= k(k + 1) + 2(k + 1) …..[From (i)]
= (k + 1).(k + 2)
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 4 + 6 + …… + 2n = n(n + 1) for all n ∈ N.

Question 2.
3 + 7 + 11 + ……… to n terms = n(2n + 1)
Solution:
Let P(n) = 3 + 7 + 11 + ……… to n terms = n(2n +1), for all n ∈ N.
But 3, 7, 11, …. are in A.P.
∴ a = 3 and d = 4
Let t_n be the nth term.
∴ t_n = a + (n – 1)d = 3 + (n – 1)4 = 4n – 1
∴ P(n) = 3 + 7 + 11 + ……. + (4n – 1) = n(2n + 1)

Step I:
Put n = 1
L.H.S. = 3
R.H.S. = 1[2(1)+ 1] = 3
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 3 + 7 + 11 + ….. + (4k – 1) = k(2k + 1) …..(i)

Sept III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
3 + 7 + 11 + …+ [4(k + 1) – 1] = (k + 1)(2k + 3)
L.H.S. = 3 + 7 + 11 + …… + [4(k + 1) – 1]
= 3 + 7 + 11 + ….. + (4k – 1) + [4(k+ 1) – 1]
= k(2k + 1) + (4k + 4 – 1) …..[From (i)]
= 2k² + k + 4k + 3
= 2k² + 2k + 3k + 3
= 2k(k + 1) + 3(k + 1)
= (k + 1) (2k + 3)
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 3 + 7 + 11 + ….. to n terms = n(2n + 1) for all n ∈ N.

Question 3.
1² + 2² + 3² +…..+ n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let P(n) = 1² + 2² + 3² +…..+ n² = \(\frac{n(n+1)(2 n+1)}{6}\) for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 1² = 1
RHS = \(\frac{1(1+1)[2(1)+1]}{6}=\frac{6}{6}\) = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1² + 2² + 3² +…+ k² = \(\frac{k(k+1)(2 k+1)}{6}\) …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1² + 2² + 3² + …+ n² = \(\frac{n(n+1)(2 n+1)}{6}\) for all n ∈ N.

Question 4.
1² + 3² + 5² + ….. + (2n – 1)² = \(\frac{n}{3}\) (2n – 1)(2n + 1)
Solution:
Let P(n) = 1² + 3² + 5²+…..+ (2n – 1)² = \(\frac{n}{3}\) (2n – 1)(2n + 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 1² = 1
R.H.S. = \(\frac{1}{3}\) [2(1) – 1][2(1) + 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1² + 3² + 5² +….+(2k – 1)² = \(\frac{k}{3}\) (2k – 1)(2k + 1) …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1² + 3² + 5² + …+ (2n – 1)² = \(\frac{n}{3}\) (2n – 1)(2n + 1) for all n ∈ N.

Question 5.
1³ + 3³ + 5³ + ….. to n terms = n² (2n² – 1)
Solution:
Let P(n) = 1³ + 3³ + 5³ + …. to n terms = n² (2n² – 1), for all n ∈ N.
But 1, 3, 5, are in A.P.
∴ a = 1, d = 2
Let tn be the nth term.
t_n = a + (n – 1) d = 1 + (n – 1) 2 = 2n – 1
∴ P(n) = 1³ + 3³ + 5³ +…..+ (2n – 1)³ = n² (2n² – 1)

Step I:
Put n = 1
L.H.S. = 1³ = 1
R.H.S. = 1² [2(1)² – 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1³ + 3³ + 5³ +…+ (2k – 1)³ = k² (2k² – 1) …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1³ + 3³ + 5³ + … to n terms = n² (2n² – 1) for all n ∈ N.

Question 6.
1.2 + 2.3 + 3.4 +… + n(n + 1) = \(\frac{n}{3}\) (n + 1)(n + 2)
Solution:
Let P(n) = 1.2 + 2.3 + 3.4 +….+n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\), for all n ∈ N.

Step I:
Put n = 1
L.H.S. = 1.2 = 2
R.H.S. = \(\frac{1}{3}\) (1 + 1)(1 + 2) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1.2 + 2.3 + 3.4 + ….. + k(k + 1) = \(\frac{k}{3}\) (k + 1)(k + 2) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1.2 + 2.3 + 3.4 + … + n(n + 1) = \(\frac{n}{3}\) (n + 1)(n + 2), for all n ∈ N.

Question 7.
1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}{3}\) (4n² + 6n – 1)
Solution:
Let P(n) = 1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}{3}\) (4n² + 6n -1), for all n ∈ N.
But first factor in each term, i.e., 1, 3, 5,… are in A.P. with a = 1 and d = 2.
∴ nth term = a + (n – 1)d = 1 + (n – 1) 2 = (2n – 1)
Also, second factor in each term,
i.e., 3, 5, 7, … are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1) d = 3 + (n – 1) 2 = (2n + 1)
∴ nth term, t_n = (2n – 1) (2n + 1)
∴ P(n) ≡ 1.3 + 3.5 + 5.7 + …. + (2n – 1) (2n + 1) = \(\frac{n}{3}\) (4n² + 6n – 1)

Step I:
Put n = 1
L.H.S. = 1.3 = 3
R.H.S. = \(\frac{1}{3}\) [4(1)² + 6(1) – 1] = 3
∴ L.H.S. = R.H.S.
∴ P(n) is trae for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1.3 + 3.5 + 5.7 +….+ (2k – 1)(2k + 1) = \(\frac{k}{3}\) (4k² + 6k – 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}{3}\) (4n² + 6n – 1) for all n ∈ N.

Question 8.
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)
Solution:
Let P(n) ≡ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\), for all n ∈ N.

Step I:
Put n = 1
L.H.S. = \(\frac{1}{1.3}=\frac{1}{3}\)
R.H.S. = \(\frac{1}{2(1)+1}=\frac{1}{3}\)
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 k-1)(2 k+1)}=\frac{k}{2 k+1}\) …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\), for all n ∈ N.

Question 9.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}\)
Solution:
Let P(n) ≡ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}\), for all n ∈ N.
But first factor in each term of the denominator,
i.e., 3, 5, 7, ….. are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1)d = 3 + (n – 1) 2 = (2n + 1)
Also, second factor in each term of the denominator,
i.e., 5, 7, 9, … are in A.P. with a = 5 and d = 2.
∴ nth term = a + (n – 1) d = 5 + (n – 1) 2 = (2n + 3)
∴ nth term, t_n = \(\frac{1}{(2 n+1)(2 n+3)}\)
P(n) ≡ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}\) = \(\frac{n}{3(2 n+3)}\)

Step I:
Put n = 1
L.H.S. = \(\frac{1}{3.5}=\frac{1}{15}\)
R.H.S. = \(\frac{1}{3[2(1)+3]}=\frac{1}{3(2+3)}=\frac{1}{15}\)
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 k+1)(2 k+3)}\) = \(\frac{k}{3(2 k+3)}\) ….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that


∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}\), for all n ∈ N.

Question 10.
(2³ⁿ – 1) is divisible by 7.
Solution:
(2³ⁿ – 1) is divisible by 7 if and only if (2³ⁿ – 1) is a multiple of 7.
Let P(n) ≡ (2³ⁿ – 1) = 7m, where m ∈ N.

Step I:
Put n = 1
∴ 2³ⁿ – 1 = 2³⁽¹⁾ – 1 = 2³ – 1 = 8 – 1 = 7
∴ (2³ⁿ – 1) is a multiple of 7.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
i.e., 2^3k – 1 is a multiple of 7.
∴ 2^3k – 1 = 7a, where a ∈ N
∴ 2^3k = 7a + 1 ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2^3(k+1) – 1 = 7b, where b ∈ N.
∴ P(k + 1) = 2^3(k+1) – 1
= 2^3k+3 – 1
= 2^3k . (2³) – 1
= (7a + 1)8 – 1 …..[From (i)]
= 56a + 8 – 1
= 56a + 7
= 7(8a + 1)
7b, where b = (8a + 1) ∈ N
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (2⁴ⁿ – 1) is divisible by 7, for all n ∈ N.

Question 11.
(2⁴ⁿ – 1) is divisible by 15.
Solution:
(2⁴ⁿ – 1) is divisible by 15 if and only if (2⁴ⁿ – 1) is a multiple of 15.
Let P(n) ≡ (2⁴ⁿ – 1) = 15m, where m ∈ N.

Step I:
Put n = 1
∴ 2⁴⁽¹⁾ – 1 = 16 – 1 = 15
∴ (2⁴ⁿ – 1) is a multiple of 15.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2^4k – 1 = 15a, where a ∈ N
∴ 2^4k = 15a + 1 …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
∴ 2^4(k+1) – 1 = 15b, where b ∈ N
∴ P(k + 1) = 2^4(k+1) – 1 = 2^4k+4 – 1
= 2^4k . 2⁴ – 1
= 16 . (2^4k) – 1
= 16(15a + 1) – 1 …..[From (i)]
= 240a + 16 – 1
= 240a + 15
= 15(16a + 1)
= 15b, where b = (16a + 1) ∈ N
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (2⁴ⁿ – 1) is divisible by 15, for all n ∈ N.

Question 12.
3ⁿ – 2n – 1 is divisible by 4.
Solution:
(3ⁿ – 2n – 1) is divisible by 4 if and only if (3ⁿ – 2n – 1) is a multiple of 4.
Let P(n) ≡ (3ⁿ – 2n – 1) = 4m, where m ∈ N.

Step I:
Put n = 1
∴ (3ⁿ – 2n – 1) = 3⁽¹⁾ – 2(1) – 1 = 0 = 4(0)
∴ (3ⁿ – 2n – 1) is a multiple of 4.
∴ P(n) is tme for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 3^k – 2k – 1 = 4a, where a ∈ N
∴ 3^k = 4a + 2k + 1 ….(i)

Step III:
We have to prove that P(n) is tme for n = k + 1,
i.e., to prove that
3^(k+1) – 2(k + 1) – 1 = 4b, where b ∈ N
P(k + 1) = 3^k+1 – 2(k + 1) – 1
= 3^k . 3 – 2k – 2 – 1
= (4a + 2k + 1) . 3 – 2k – 3 …….[From (i)]
= 12a + 6k + 3 – 2k – 3
= 12a + 4k
= 4(3a + k)
= 4b, where b = (3a + k) ∈ N
∴ P(n) is tme for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is tme for all n ∈ N.
∴ 3ⁿ – 2n – 1 is divisible by 4, for all n ∈ N.

Question 13.
5 + 5² + 5³ + ….. + 5ⁿ = \(\frac{5}{4}\) (5ⁿ – 1)
Solution:
Let P(n) ≡ 5 + 52 + 53 +…..+ 5n = \(\frac{5}{4}\) (5n – 1), for all n ∈ N.

Step I:
Put n = 1
L.H.S. = 5
R.H.S. = \(\frac{5}{4}\) (5¹ – 1) = 5
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 5 + 5² + 5³ + ….. + 5^k = \(\frac{5}{4}\) (5^k – 1) …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 5 + 5² + 5³ + … + 5ⁿ = \(\frac{5}{4}\) (5ⁿ – 1), for all n ∈ N.

Question 14.
(cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ)
Solution:
Let P(n) ≡ (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, for all n ∈ N.
Step I:
Put n = 1
L.H.S. = (cos θ + i sin θ)¹ = cos θ + i sin θ
R.H.S. = cos[(1)θ] + i sin[(1)θ] = cos θ + i sin θ
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ (cos θ + i sin θ)^k = cos kθ + i sin kθ …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
(cos θ + i sin θ)^k+1 = cos (k + 1)θ + i sin (k + 1)θ
L.H.S. = (cos θ + i sin θ)^k+1
= (cos θ + i sin θ)^k . (cos θ + i sin θ)
= (cos kθ + i sin kθ) . (cos θ + i sin θ) ……[From (i)]
= cos kθ cos θ + i sin θ cos kθ + i sin kθ cosθ – sin kθ sin θ ……[∵ i² = -1]
= (cos kθ cos θ – sin k θ sin θ) + i(sin kθ cos θ + cos kθ sin θ)
= cos(kθ + θ) + i sin(kθ + θ)
= cos(k + 1) θ + i sin (k + 1) θ
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ), for all n ∈ N.

Question 15.
Given that t_n+1 = 5 t_n+4, t₁ = 4, prove by method of induction that t_n = 5ⁿ – 1.
Solution:
Let the statement P(n) has L.H.S. a recurrence relation t_n+1 = 5 t_n+4, t₁ = 4 and R.H.S. a general statement t_n = 5ⁿ – 1.
Step I:
Put n = 1
L.H.S. = 4
R.H.S. = 5¹ – 1 = 4
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S. = t₂ = 5t₁ + 4 = 24
R.H.S. = t₂ = 5² – 1 = 24
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 2.

Step II:
Let us assume that P(n) is true for n = k.
∴ t_k+1 = 5 t_k+4 and t_k = 5^k – 1

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that t_k+1 = 5^k+1 – 1
Since t_k+1 = 5 t_k+4 and t_k = 5^k – 1 …..[From Step II]
t_k+1 = 5 (5^k – 1) + 4 = 5^k+1 – 1
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ t_n = 5ⁿ – 1, for all n ∈ N.

Question 16.
Prove by method of induction
\(\left(\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right)^{n}=\left(\begin{array}{cc}
1 & 2 n \\
0 & 1
\end{array}\right) \forall n \in N\)
Solution:

Class 11 Maharashtra State Board Maths Solution 

• Exercise 4.1 Class 11 Maths 2
• Exercise 4.2 Class 11 Maths 2
• Exercise 4.3 Class 11 Maths 2
• Exercise 4.4 Class 11 Maths 2
• Exercise 4.5 Class 11 Maths 2
• Miscellaneous Exercise 4 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Miscellaneous Exercise 3 Questions and Answers.

11th Maths Part 2 Permutations and Combination Miscellaneous Exercise 3 Questions And Answers Maharashtra Board

(I) Select the correct answer from the given alternatives.

Question 1.
A college offers 5 courses in the morning and 3 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening is
(A) 5
(B) 3
(C) 8
(D) 15
Answer:
(C) 8
Hint:
Number of ways to select one course from available 8 courses
(i.e., 5 courses in the morning and 3 in the evening) = 5 + 3 = 8

Question 2.
A college has 7 courses in the morning and 3 in the evening. The possible number of choices with the student if he wants to study one course in the morning and one in the evening is
(A) 21
(B) 4
(C) 42
(D) 10
Answer:
(A) 21
Hint:
Number of ways to select one morning and one evening course = ⁷C₁ × ³C₁ = 21

Question 3.
In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all persons of the same nationality sit together?
(A) 3! 8!
(B) 3! 4! 8! 4!
(C) 4! 4!
(D) 8! 4! 4!
Answer:
(B) 3! 4! 8! 4!
Hint:
8 Indians take their seats in 8! ways, 4 Americans take their seats in 4! ways, 4 Englishmen take their seats in 4! ways.
Three groups of Indians, Americans and Englishmen can be permuted in 3! ways.
Required number = 3! × 8! × 4! × 4!

Question 4.
In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?
(A) 9 × 8!
(B) 8 × 8!
(C) 9 × 9!
(D) 8 × 9!
Answer:
(D) 8 × 9!
Hint:
Arrange 8 papers in 8! ways and two papers in 9 gaps are arranged in ⁹P₂ ways.
Required number = 8! ⁹P₂
= 8! × 9 × 8
= 9! × 8

Question 5.
In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.
(A) 12
(B) 288
(C) 144
(D) 256
Answer:
(C) 144
Hint:
B G B G B G B
4 boys take their seats in 4! ways.
3 girls take their seats in 3! ways.
Required number = 4! × 3!
= 24 × 6
= 144

Question 6.
Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.
(A) 16
(B) 56
(C) 24
(D) 8
Answer:
(B) 56
Hint:
A triangle is obtained by joining three vertices.
Number of ways of selecting 3 vertices out of 8 vertices = ⁸C₃
= \(\frac{8 \times 7 \times 6}{1 \times 2 \times 3}\)
= 56

Question 7.
A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?
(A) 320
(B) 750
(C) 40
(D) 11340
Answer:
(D) 11340
Hint:
Number of ways to choose 8 questions from Part A and 5 from Part B = ¹⁰C₈ × ¹⁰C₅
= ¹⁰C₂ × ¹⁰C₅
= 45 × 252
= 11340

Question 8.
There are 10 persons among whom two are brothers. The total number of ways in which these persons can be seated around a round table so that exactly one person sits between the brothers is equal to:
(A) 2! × 7!
(B) 2! × 8!
(C) 3! × 7!
(D) 3! × 8!
Answer:
(B) 2! × 8!
Hint:
Select a person from 8 people (i.e., the people excluding two brothers).
This is done in 8 ways.
2 brothers sit adjacent to the selected person on two sides, they may interchange their seats.
Remaining 7 people sit in 7! ways
Required number = 8 × 2 × 7! = 2! × 8!

Question 9.
The number of arrangements of the letters of the word BANANA in which two N’s do not appear adjacently is
(A) 80
(B) 60
(C) 40
(D) 100
Answer:
(C) 40
Hint:
Arrange B, A, A, A in \(\frac{4 !}{3 !}\) ways.
These four letters create 5 gaps in which 2 N are to be filled, this can be done in 5C2 ways, we do not permute those 2N as they are identical.
∴ Required number = \(\frac{4 !}{3 !}\) × ⁵C₂ = 40

Question 10.
The number of ways in which 5 male and 2 female members of a committee can be seated around a round table so that the two females are not seated together is
(A) 840
(B) 600
(C) 720
(D) 480
Answer:
(D) 480
Hint:
5 males take their seats in 4! ways, creating 5 gaps.
In these 5 gaps, 2 females are to be seated.
∴ The number of ways to do this = ⁵C₂ × 2!
Required number = 4! × ⁵C₂ × 2! = 480

(II) Answer the following.

Question 1.
Find the value of r if ⁵⁶P_r+2 : ⁵⁴P_r-1 = 30800 : 1.
Solution:

Question 2.
How many words can be formed by writing letters in the word CROWN in a different order?
Solution:
Five Letters of the word CROWN are to be permuted.
∴ Number of different words = 5! = 120

Question 3.
Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Solution:
There are 6 letters A, E, I, M, N, R.
Number of words that can be formed by using all these letters = 6! = 720
When a word starts with ‘A’,
‘A’ can be arranged in 1 way and the remaining 5 letters can be arranged among themselves in 5! ways.
The number of words starting with A = 5!
∴ Similarly,
The number of words starting with E = 5!
The number of words starting with I = 5!
The number of words starting with M = 5!
The number of words starting with N = 5!
The number of words starting with R = 5!
Total number of words = 6 × 5! = 720
Number of words starting with AE = 4! = 24
Number of words starting with AIE = 3! = 6
Number of words starting with AIM = 3! = 6
Number of words starting with AINE = 2!
Total words = 24 + 6 + 6 + 2 = 38
39th word is AINMER
40th word is AINMRE

Question 4.
The Capital English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letters passwords can be formed using these letters?
Solution:
There are 11 symmetric letters.
∴ Number of 3 Letter passwords = ¹¹P₃
= 11 × 10 × 9
= 990

Question 5.
How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?
Solution:
A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also, among the given numbers 2 is repeated twice and 3 is repeated thrice.
∴ Required number of numbers = Total number of arrangements possible among these digits – number of arrangements of 7 digits which begin with 0.
= \(\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 3 !}-\frac{6 \times 5 \times 4 \times 3 !}{2 \times 3 !}\)
= 7 × 6 × 5 × 2 – 6 × 5 × 2
= 6 × 5 × 2(7 – 1)
= 60 × 6
= 360

Question 6.
Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Solution:
Ten students are to be selected for a project from a class of 30 students.
Case I:
If 4 students join the project, then from remaining 26 students, rest of the 6 students are to be selected.
Which can be done in ²⁶C₆
= \(\frac{26 !}{6 !(26-6) !}\)
= \(\frac{26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 !}{6 ! \times 20 !}\)
= 230230 ways.

Case II:
If 4 students does not join the project, then from remaining 26 students, all the 10 students are to be selected.
Which can be done in ²⁶C₁₀
= \(\frac{26 !}{10 !(26-10) !}\)
= \(\frac{26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 !}{10 ! \times 16 !}\)
= 5311735 ways.
∴ Required number of selections = ²⁶C₆ + ²⁶C₁₀
= 230230 + 5311735
= 5541965

Question 7.
A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Solution:
There are 7 books of student’s interest, but he can borrow only three books.
He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed.
Consider the following table:

Required number of selections = 5 + 10 = 15

Question 8.
30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so.
Solution:
First we can select 7 objects out of 30 for the first group in ³⁰C₇ ways.
Now there are 23 objects left out of which we can select 10 objects for the second group in ²³C₁₀ ways.
Remaining 13 objects can be selected for the third group in ⁵C₅ ways.
∴ Required number of ways = ³⁰C₇ × ²³C₁₀ × ¹³C₁₃
= \(\frac{30 !}{23 ! 7 !} \times \frac{23 !}{10 ! 13 !} \times 1\)
= \(\frac{30 !}{7 ! 10 ! 13 !}\)

Question 9.
A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
Solution:
Every subject a student may pass or fail.
∴ Total number of outcomes = 2⁷ = 128
This number includes one case when the student passes in all subjects.
Required number of ways = 128 – 1 = 127

Question 10.
Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Solution:
Nine friends decide to go for a picnic in two groups and there must be at least 3 friends in each group.
Consider the following table:

Question 11.
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
Solution:
Every lamp is either ON or OFF.
There are 12 lamps
Number of instances = 2¹²
This number includes one case in when all 12 lamps are OFF.
∴ Required Number of ways = 2¹² – 1 = 4095

Question 12.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Solution:
A quadratic equation is to be formed using numbers 0, 2, 4, 5 as coefficients and a coefficient can be repeated.
Let the quadratic equation be ax² + bx + c = 0, a ≠ 0
Consider the following table:

Number of quadratic equations can be formed = 3 × 4 × 4 = 48

Question 13.
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Solution:
There are total of 10 digits.
Let the telephone number be 45abcd.
There are 8 digits left for the choice of a, b, c, d as repetition is not allowed.
Consider the following table:

∴ Required number of numbers formed = 8 × 7 × 6 × 5 = 1680

Question 14.
A question paper has 6 questions. How many ways does a student have to answer if he wants to solve at least one question?
Solution:
Every question is ‘SOLVED’ or ‘NOT SOLVED’.
There are 6 questions.
Number of outcomes = 2⁶
This number includes one case when the student solves NONE of the questions.
∴ Required number of ways = 2⁶ – 1 = 64 – 1 = 63

Question 15.
Find the number of ways of dividing 20 objects into three groups of sizes 8, 7, and 5.
Solution:
First we can select 8 objects our of 20 for the first group in ²⁰C₈ ways.
Now there are 12 objects left out of which we can select 7 objects for the second group in ¹²C₇ ways.
Remaining 5 objects can be selected for the third group in ⁵C₅ ways.
∴ Required number of ways = ²⁰C₈ × ¹²C₇ × ⁵C₅
= \(\frac{20 !}{8 ! 12 !} \times \frac{12 !}{7 ! 5 !} \times 1\)
= \(\frac{20 !}{8 ! 7 ! 5 !}\)

Question 16.
There are 4 doctors and 8 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
Solution:
There are 4 doctors and 8 lawyers in a panel.
A team of 6 with at least one doctor is to be formed.
We count the number by the INDIRECT method of counting.
Number of ways to select a team of 6 people = ¹²C₆
Number of teams with No doctor in any team = ⁸C₆
∴ Required number of ways = ¹²C₆ – ⁸C₆
= 924 – 28
= 896

Question 17.
Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms formed.
Solution:
The first set has 4 parallel lines and another set has 5 parallel lines.
To form a parallelogram, we need 2 lines from each set.
∴ Required number of distinct parallelograms formed = ⁴C₂ × ⁵C₂
= 6 × 10
= 60

Question 18.
There are 12 distinct points A, B, C, …, L, in order, on a circle. Lines are drawn passing through each pair of points.
(i) How many lines are there in total?
(ii) How many lines pass through D?
(iii) How many triangles are determined by lines?
(iv) How many triangles have on vertex C?
Solution:
(i) We need two points to draw a line.
∴ Total number of lines = ¹²C₂ = 66

(ii) Lines are drawn passing through each pair of points.
∴ Lines from point D will pass through all the remaining 11 points.
∴ 11 lines pass through D.

(iii) We need three points to draw a triangle.
∴ Number of triangles = ¹²C₃ = 220

(iv) To get the triangles with one vertex as C,
we need two vertices from the remaining 11 vertices.
∴ Number of triangles with vertex at C = ¹¹C₂
= \(\frac{11 \times 10}{2}\)
= 55

Class 11 Maharashtra State Board Maths Solution 

• Exercise 3.1 Class 11 Maths 2
• Exercise 3.2 Class 11 Maths 2
• Exercise 3.3 Class 11 Maths 2
• Exercise 3.4 Class 11 Maths 2
• Exercise 3.5 Class 11 Maths 2
• Exercise 3.6 Class 11 Maths 2
• Miscellaneous Exercise 3 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.6 Questions and Answers.

11th Maths Part 2 Permutations and Combination Exercise 3.6 Questions And Answers Maharashtra Board

Question 1.
Find the value of
(a) ¹⁵C₄
Solution:

(b) ⁸⁰C₂
Solution:

(c) ¹⁵C₄ + ¹⁵C₅
Solution:

(d) ²⁰C₁₆ – ¹⁹C₁₆
Solution:

Question 2.
Find n if
(a) ⁶P₂ = n(⁶C₂)
Solution:

(b) ²ⁿC₃ : ⁿC₂ = 52 : 3
Solution:

(c) ⁿC_n-3 = 84
Solution:

Question 3.
Find r if ¹⁴C_2r : ¹⁰C_2r-4 = 143 : 10.
Solution:

∴ 2r(2r – 1) (2r – 2) (2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1) (2r – 2) (2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4

Question 4.
Find n and r if,
(a) ⁿP_r = 720 and ⁿC_n-r = 120
Solution:

(b) ⁿC_r-1 : ⁿC_r : ⁿC_r+1 = 20 : 35 : 42
Solution:

Question 5.
If ⁿP_r = 1814400 and ⁿC_r = 45, find ⁿ⁺⁴C_r+3.
Solution:

Question 6.
If ⁿC_r-1 = 6435, ⁿC_r = 5005, ⁿC_r+1 = 3003, find ^rC₅.
Solution:

Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 8 green balls, and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 8 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in ⁶C₃ ways.
3 reen balls can be selected from 8 green balls in ⁸C₃ ways.
3 blue balls can be selected from 7 blue balls in ⁷C₃ ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in ⁶C₃ ways.
2 girls can be selected from 4 girls in ⁴C₂ ways.
∴ Number of ways the team can be selected

Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = ⁿC₂
Given 66 handshakes were exchanged.
66 = ⁿC₂
66 = \(\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}\)
66 × 2 = \(\frac{n(n-1)(n-2) !}{(n-2) !}\)
132 = n (n – 1)
n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.

Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points

Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
(iv) n = 8
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides.
∴ Through ‘n’ points we can draw ⁿC₂ lines including sides.
∴ Number of diagonals in n sided polygon = ⁿC₂ – n (n = number of sides)

Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel, they intersect at a point.
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if (a) no three points are collinear (b) four points are collinear
Solution:
There are 10 points on a plane.
(a) When no three of them are collinear.
A line is obtained by joining 2 points.
∴ Number of lines passing through these points = ¹⁰C₂
= \(\frac{10 !}{2 ! 8 !}\)
= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\)
= 5 × 9
= 45

(b) When 4 of them are collinear.
If no three points are collinear, we get a total of ¹⁰C₂ = 45 lines by joining them. …..[From (i)]
Since 4 points are collinear, only one line passes through these points instead of ⁴C₂ lines.
∴ ⁴C₂ – 1 extra lines are included in 45 lines.
Number of lines passing through these points
= 45 – (⁴C₂ – 1)
= 45 – \(\frac{4 !}{2 ! 2 !}\) + 1
= 45 – \(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\) + 1
= 45 – 6 + 1
= 40

Question 14.
Find the number of triangles formed by joining 12 points if
(a) no three points are collinear
(b) four points are collinear
Solution:
There are 12 points on the plane.
(a) When no three of them are collinear.
A triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = ¹²C₃
= \(\frac{12 !}{3 ! 9 !}\)
= \(\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}\)
= 220

(b) When 4 of these points are collinear.
If no three points are collinear, total we get ¹²C₃ = 220 triangles by joining them. ……[From (i)]
Since 4 points are collinear, no triangle can be formed by joining these four points.
∴ ⁴C₃ extra triangles are included in 220 triangles.
∴ Number of triangles that can be obtained from these points = ¹²C₃ – ⁴C₃
= 220 – \(\frac{4 !}{3 ! \times 1 !}\)
= 220 – \(\frac{4 \times 3 !}{3 !}\)
= 220 – 4
= 216

Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
There are 8 consonants and 3 vowels.
From 8 consonants, 4 can be selected in ⁸C₄
= \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}\)
= 70 ways.
From 3 vowels, 2 can be selected in ³C₂
= \(\frac{3 !}{2 ! 1 !}\)
= \(\frac{3 \times 2 !}{2 !}\)
= 3 ways.
Now, to form a word, these 6 ietters (i.e., 4 consonants and 2 vowels) can be arranged in ⁶P₆ = 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.

Question 16.
Find n if,
(i) ⁿC₈ = ⁿC₁₂
Solution:
ⁿC₈ = ⁿC₁₂
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 8 = 12 or 8 = n – 12
But 8 = 12 is not possible
∴ 8 = n – 12
∴ n = 20

(ii) ²³C_3n = ²³C_2n+3
Solution:
²³C_3n = ²³C_2n+3
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 3n = 2n + 3 or 3n = 23 – 2n – 3
∴ n = 3 or n = 4

(iii) ²¹C_6n = \({ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}\)
Solution:
²¹C_6n = \({ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}\)
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 6n = n² + 5 or 6n = 21 – (n² + 5)
∴ n² – 6n + 5 = 0 or 6n = 21 – n² – 5
∴ n² – 6n + 5 = 0 or n² + 6n – 16 = 0
If n² – 6n + 5 = 0, then (n – 1)(n – 5) = 0
∴ n = 1 or n = 5
If n = 5 then
n² + 5 = 30 > 21
∴ n ≠ 5
∴ n = 1
If n² + 6n – 16 = 0, then (n + 8)(n – 2) = 0
n = -8 or n = 2
n ≠ -8
∴ n = 2
∴ n = 1 or n = 2

Check:
n = 2
∴ n² + 5 = 2² + 5 = 9
²¹C_6n = ²¹C₁₂
and \({ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}\) = ²¹C₉
∴ ⁿC_r = ⁿC_n-r
∴ ²¹C₁₂ = ²¹C₉
∴ n = 2 is a right answer.

(iv) ²ⁿC_r-1 = ²ⁿC_r+1
Solution:
²ⁿC_r-1 = ²ⁿC_r+1
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ r – 1 = r + 1 or r – 1 = 2n – (r + 1)
But r – 1 = r + 1 is not possible
∴ r – 1 = 2n – (r + 1)
∴ r + r = 2n
∴ r = n

Check:
²ⁿC_r-1 = ²ⁿC_n-1
and ²ⁿC_r+1 = ²ⁿC_n+1
using ⁿC_r = ⁿC_n-r, we have
²ⁿC_n+1 = ²ⁿC_2n-(n+1) = ²ⁿC_n-1
∴ ²ⁿC_r-1 = ²ⁿC_r+1

(v) ⁿC_n-2 = 15
Solution:
ⁿC_n-2 = 15
∴ ⁿC₂ = 15 …..[∵ ⁿC_r = ⁿC_n-r]
∴ \(\frac{n !}{(n-2) ! 2 !}=15\)
∴ \(\frac{n(n-1)(n-2) !}{(n-2) ! \times 2 \times 1}=15\)
∴ n(n – 1) = 30
∴ n(n – 1) = 6 × 5
Equating both sides, we get
∴ n = 6

Question 17.
Find x if ⁿP_r = x ⁿC_r.
Solution:

Question 18.
Find r if ¹¹C₄ + ¹¹C₅ + ¹²C₆ + ¹³C₇ = ¹⁴C_r.
Solution:

Question 19.
Find the value of \(\sum_{r=1}^{4}{ }^{(21-r)} \mathrm{C}_{4}\).
Solution:

Question 20.
Find the differences between the greatest values in the following:
(a) ¹⁴C_r and ¹²C_r
Solution:
Greatest value of ¹⁴C_r.
Here, n = 14, which is even.
Greatest value of nCr occurs at r = \(\frac{n}{2}\) if n is even.
∴ r = \(\frac{n}{2}\)
∴ r = \(\frac{14}{2}\) = 7

∴ Difference between the greatest values of ¹⁴C_r and ¹²C_r = ¹⁴C_r – ¹²C_r
= 3432 – 924
= 2508

(b) ¹³C_r and ⁸C_r
Solution:
Greatest value of ¹³C_r.
Here n = 13, which is odd.
Greatest value of nCr occurs at r = \(\frac{n-1}{2}\) if n is odd.
∴ r = \(\frac{\mathrm{n}-1}{2}\)
∴ r = \(\frac{13-1}{2}\) = 6

∴ Difference between the greatest values of ¹³C_r and ⁸C_r = ¹³C_r – ⁸C_r
= 1716 – 70
= 1646

(c) ¹⁵C_r and ¹¹C_r
Solution:
Greatest value of ¹⁵C_r.
Here n = 15, which is odd.
Greatest value of nCr occurs at r = \(\frac{n-1}{2}\) if n is odd.
∴ r = \(\frac{n-1}{2}\)
∴ r = \(\frac{15-1}{2}\) = 7

∴ Difference between the greatest values of ¹⁵C_r and ¹¹C_r = ¹⁵C_r – ¹¹C_r
= 6435 – 462
= 5973

Question 21.
In how many ways can a boy invite his 5 friends to a party so that at least three join the party?
Solution:
Boy can invite = (3 or 4 or 5 friends)
Consider the following table:

∴ Number of ways a boy can invite his friends to a party so that three or more of join the party = 10 + 5 + 1 = 16

Question 22.
A group consists of 9 men and 6 women. A team of 6 is to be selected. How many of possible selections will have at least 3 women?
Solution:
There are 9 men and 6 women.
A team of 6 persons is to be formed such that it consist of at least 3 women.
Consider the following table:

∴ No. of ways this can be done = 1680 + 540 + 54 + 1 = 2275
∴ 2275 teams can be formed if team consists of at least 3 women.

Question 23.
A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in majority?
Solution:
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women.
Consider the following table:

∴ Number of committees with at least 5 women
= 14112 + 14700 + 6720 + 1260 + 81
= 36873

(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)
= ¹⁸C₁₀ – 36873
= ¹⁸C₈ – 36873
= 43758 – 36873
= 6885

Question 24.
A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
Solution:
There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.
The student has to select 6 questions taking at least 2 questions from each section.
Consider the following table:

∴ Number of choices = 150 + 200 + 75 = 425
∴ In 425 ways students can select 6 questions, taking at least 2 questions from each section.

Question 25.
There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 player is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
Solution:
There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.
A team of 11 players is to be chosen such that exactly one wicket keeper and at least 4 bowlers are to be included in the team.
Consider the following table:

∴ Number of ways a team of 11 players can be selected
= 45045 + 6006
= 51051

Question 26.
Five students are selected from 11. How many ways can these students be selected if
(a) two specified students are selected?
(b) two specified students are not selected?
Solution:
5 students are to be selected from 11 students.
(a) When 2 specified students are included,
then remaining 3 students can be selected from (11 – 2) = 9 students.
∴ Number of ways of selecting 3 students from 9 students = ⁹C₃
= \(\frac{9 !}{3 ! \times 6 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}\)
= 84
∴ Selection of students is done in 84 ways when 2 specified students are included.

(b) When 2 specified students are not included, then 5 students can be selected from the remaining (11 – 2) = 9 students.
∴ Number of ways of selecting 5 students from 9 students = ⁹C₅
= \(\frac{9 !}{5 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1}\)
= 126
∴ Selection of students is done in 126 ways when 2 specified students are not included.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 3.1 Class 11 Maths 2
• Exercise 3.2 Class 11 Maths 2
• Exercise 3.3 Class 11 Maths 2
• Exercise 3.4 Class 11 Maths 2
• Exercise 3.5 Class 11 Maths 2
• Exercise 3.6 Class 11 Maths 2
• Miscellaneous Exercise 3 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.5 Questions and Answers.

11th Maths Part 2 Permutations and Combination Exercise 3.5 Questions And Answers Maharashtra Board

Question 1.
In how many different ways can 8 friends sit around a table?
Solution:
We know that ‘n’ persons can sit around a table in (n – 1)! ways.
∴ 8 friends can sit around a table in 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.

Question 2.
A party has 20 participants. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have two specified persons on either side of the host?
Solution:
A party has 20 participants.
All of them and the host (i.e., 21 persons) can be seated at a circular table in (21 – 1)! = 20! ways.
When two particular participants are seated on either side of the host.
The host takes the chair in 1 way.
These 2 persons can sit on either side of the host in 2! ways.
Once the host occupies his chair, it is not circular permutation more.
The remaining 18 people occupy their chairs in 18! ways.
∴ A total number of arrangements possible if two particular participants are seated on either side of the host = 2! × 18! = 2 × 18!

Question 3.
Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are (a) always together. (b) never together.
Solution:
(a) Delegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.
Let us consider these 2 delegates as one unit. They can be arranged among themselves in 2! ways.
Also, these two delegates are to be seated with 22 other delegates (i.e. total of 23) which can be done in (23 – 1)! = 22! ways.
∴ Required number of arrangements = 2! × 22!

(b) When 2 specified delegates are never together then, other 22 delegates can be participate in a round table discussion in (22 – 1)! = 21! ways.
∴ There are 22 places of which any 2 places can be filled by those 2 delegates so that they are never together.
∴ Two specified delegates can be arranged in ²²P₂ ways.
∴ Required number of arrangements = ²²P₂ × 21!
= \(\frac{22 !}{(22-2) !} \times 21 !\)
= \(\frac{22 !}{20 !}\) × 21!
= 22 × 21 × 21!
= 21 × 22 × 21!
= 21 × 22!

Question 4.
Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbours.
Solution:
There are 15 people to sit around a table.
∴ They can be arranged in(15 – 1)! = 14! ways.
But, they should not have the same neighbour in any two arrangements.
Around the table, arrangements (i.e., clockwise and anticlockwise) coincide.

∴ Required number of arrangements = \(\frac{14 !}{2}\)

Question 5.
A committee of 10 members sits around a table. Find the number of arrangements that have the President and the Vice-president together.
Solution:
A committee of 10 members sits around a table.
But, President and Vice-president sit together.
Let us consider President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.
Now, this unit with the other 8 members of the committee is to be arranged around a table, which can be done in (9 – 1)! = 8! ways.
∴ Required number of arrangements = 8! × 2! = 2 × 8!

Question 6.
Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated (a) between the two women. (b) between two men.
Solution:
5 men, 2 women, and a child sit around a table.
(a) When the child is seated between two women.
5 men, 2 women, and a child are to be seated around a round table such that the child is seated between two women.
∴ the two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.
Now, this one unit is to be arranged with the remaining 5 men,
i.e., a total of 6 units are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.
∴ Required number of arrangements = 5! × 2!
= 120 × 2
= 240

(b) Two men can be selected from 5 men in
⁵C₂ = \(\frac{5 !}{2 !(5-2) !}=\frac{5 \times 4 \times 3 !}{2 \times 3 !}\) = 10 ways.
Also, these two men can sit on either side of the child in 2! ways.
Let us take two men and a child as one unit.
Now, this one unit is to be arranged with the remaining 3 men and 2 women,
i.e., a total of 6 units (3 + 2 + 1) are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.
∴ Required number of arrangements = 10 × 2! × 5!
= 10 × 2 × 120
= 2400

Question 7.
Eight men and six women sit around a table. How many sitting arrangements will have no two women together?
Solution:
8 men can be seated around a table in (8 – 1)! = 7! ways.
No two women should sit together.
There are 8 gaps created by 8 men’s seats.
∴ Women can be seated in 8 gaps in ⁸P₆ ways.
∴ Required number of arrangements = 7! × ⁸P₆

Question 8.
Find the number of seating arrangements for 3 men and 3 women to sit around a table so that exactly two women are together.
Solution:
2 women (who wish to sit together) can be selected from 3 in
³C₂ = \(\frac{3 !}{2 !(3-2) !}=\frac{3 \times 2 !}{2 ! \times 1 !}\) = 3 ways.
Also, these two women can sit together in 2! ways.
Let us take two women as one unit.
Now, this one unit is to be arranged with the remaining 3 men and 1 woman,
i.e., a total of 5 units are to be arranged around a round table, which can be done in (5 – 1)! = 4! ways.
∴ Required number of arrangements = 3 × 2! × 4!
= 3 × 2 × 24
= 144

Question 9.
Four objects in a set of ten objects are alike. Find the number of ways of arranging them in a circular order.
Solution:
Ten things can be arranged in a circular order of which 4 are alike in \(\frac{9 !}{4 !}\) ways.
∴ Required number of arrangements = \(\frac{9 !}{4 !}\)

Question 10.
Fifteen persons sit around a table. Find the number of arrangements that have two specified persons not sitting side by side.
Solution:
Since 2 particular persons can’t be sitting side by side,
the other 13 persons can be arranged around the table in (13 – 1)! = 12! ways.
The two persons who are not sitting side by side may take 13 positions created by 3 persons in ¹³P₂ ways.
∴ Required number of arrangements = 12! × ¹³P₂
= 12! × 13 × 12
= 13 × 12! × 12
= 12 × 13!

Class 11 Maharashtra State Board Maths Solution 

• Exercise 3.1 Class 11 Maths 2
• Exercise 3.2 Class 11 Maths 2
• Exercise 3.3 Class 11 Maths 2
• Exercise 3.4 Class 11 Maths 2
• Exercise 3.5 Class 11 Maths 2
• Exercise 3.6 Class 11 Maths 2
• Miscellaneous Exercise 3 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.4 Questions and Answers.

11th Maths Part 2 Permutations and Combination Exercise 3.4 Questions And Answers Maharashtra Board

Question 1.
Find the number of permutations of letters in each of the following words.
(i) DIVYA
(ii) SHANTARAM
(iii) REPRESENT
(iv) COMBINE
(v) BAL BHARATI
Solution:
(i) There are 5 distinct letters in the word DIVYA.
∴ Number of permutations of the letters of the word DIVYA = 5! = 120

(ii) There are 9 letters in the word SHANTARAM in which ‘A’ is repeated 3 times.
∴ Number of permutations of the letters of the word SHANTARAM = \(\frac{9 !}{3 !}\)
= 9 × 8 × 7 × 6 × 5 × 4
= 60480

(iii) There are 9 letters in the word REPRESENT in which ‘E’ is repeated 3 times and ‘R’ is repeated 2 times.
∴ Number of permutations of the letters of the word REPRESENT = \(\frac{9 !}{3 ! 2 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}\)
= 30240

(iv) There are 7 distinct letters in the word COMBINE.
∴ Number of permutations of the letters of the word COMBINE = 7! = 5040

(v) There are 10 letters in the word BALBHARATI in which ‘B’ is repeated 2 times and ‘A’ is repeated 3 times.
∴ Number of permutations of the letters of the word BALBHARATI = \(\frac{10 !}{2 ! 3 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 3 \times 2}\)
= 302400

Question 2.
You have 2 identical books on English, 3 identical books on Hindi, and 4 identical books on Mathematics. Find the number of distinct ways of arranging them on a shelf.
Solution:
There are total 9 books to be arranged on a shelf.
Out of these 9 books, 2 books on English, 3 books on Hindi and 4 books on mathematics are identical.
∴ Total number of arrangements possible = \(\frac{9 !}{2 ! 3 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 3 \times 2 \times 4 !}\)
= 9 × 4 × 7 × 5
= 1260

Question 3.
A coin is tossed 8 times. In how many ways can we obtain (a) 4 heads and 4 tails? (b) at least 6 heads?
Solution:
A coin is tossed 8 times. All heads are identical and all tails are identical.
(a) 4 heads and 4 tails are to be obtained.
∴ Number of ways it can be obtained = \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2}\)
= 70

(b) At least 6 heads are to be obtained.
∴ Outcome can be (6 heads and 2 tails) or (7 heads and 1 tail) or (8 heads)
∴ Number of ways it can be obtained = \(\frac{8 !}{6 ! 2 !}+\frac{8 !}{7 ! 1 !}+\frac{8 !}{8 !}\)
= \(\frac{8 \times 7}{2}\) + 8 + 1
= 28 + 8 + 1
= 37

Question 4.
A bag has 5 red, 4 blue, and 4 green marbles. If all are drawn one by one and their colours are recorded, how many different arrangements can be found?
Solution:
There is a total of 13 marbles in a bag.
Out of these 5 are Red, 4 Blue, and 4 are Green marbles.
All balls of the same colour are taken to be identical.
∴ Required number of arrangements = \(\frac{13 !}{5 ! 4 ! 4 !}\)

Question 5.
Find the number of ways of arranging letters of the word MATHEMATICAL. How many of these arrangements have all vowels together?
Solution:
There are 12 letters in the word MATHEMATICAL in which ‘M’ is repeated 2 times, ‘A’ repeated 3 times and ‘T’ repeated 2 times.
∴ Required number of arrangements = \(\frac{12 !}{2 ! 3 ! 2 !}\)
When all the vowels,
i.e., ‘A’, ‘A’, ‘A’, ‘E’, ‘I’ are to be kept together.
Let us consider them as one unit.
Number of arrangements of these vowels among themselves = \(\frac{5 !}{3 !}\) ways.
This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each.
∴ Number of such arrangements = \(\frac{8 !}{2 ! 2 !}\)
∴ Required number of arrangements = \(\frac{8 ! \times 5 !}{2 ! 2 ! 3 !}\)

Question 6.
Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have (a) letters R and H never together? (b) all vowels together?
Solution:
There are 11 letters in the word MAHARASHTRA in which ‘A’ is repeated 4 times, ‘H’ repeated 2 times, and ‘R’ repeated 2 times.
∴ Total number of words can be formed = \(\frac{11 !}{4 ! 2 ! 2 !}\)

(a) When letters R and H are never together.
Other than 2R, 2H there are 4A, 1S, 1T, 1M.
These letters can be arranged in \(\frac{7 !}{4 !}\) ways = 210.
These seven letters create 8 gaps in which 2R, 2H are to be arranged.
Number of ways to do = \(\frac{{ }^{8} \mathrm{P}_{4}}{2 ! 2 !}\) = 420
Required number of arrangements = 210 × 420 = 88200.

(b) When all vowels are together.
There are 4 vowels in the word MAHARASHTRA, i.e., A, A, A, A.
Let us consider these 4 vowels as one unit, which can be arranged among themselves in \(\frac{4 !}{4 !}\) = 1 way.
This unit is to be arranged with 7 other letters in which ‘H’ is repeated 2 times, ‘R’ is repeated 2 times.
∴ Total number of arrangements = \(\frac{8 !}{2 ! 2 !}\)

Question 7.
How many different words are formed if the letter R is used thrice and letters S and T are used twice each?
Solution:
To find the number of different words when ‘R’ is taken thrice, ‘S’ is taken twice and ‘T’ is taken twice.
∴ Total number of letters available = 7, of which ‘S’ and ‘T’ repeat 2 times each, ‘R’ repeats 3 times.
∴ Required number of words = \(\frac{7 !}{2 ! 2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 1 \times 2 \times 1 \times 3 !}\)
= 7 × 6 × 5
= 210

Question 8.
Find the number of arrangements of letters in the word MUMBAI so that the letter B is always next to A.
Solution:
There are 6 letters in the word MUMBAI.
These letters are to be arranged in such a way that ‘B’ is always next to ‘A’.
Let us consider AB as one unit.
This unit with the other 4 letters in which ‘M’ repeats twice is to be arranged.
∴ Required number of arrangements = \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60

Question 9.
Find the number of arrangements of letters in the word CONSTITUTION that begin and end with N.
Solution:
There are 12 letters in the word CONSTITUTION, in which ‘O’, ‘N’, ‘I’ repeat two times each, ‘T’ repeats 3 times.
When the arrangement starts and ends with ‘N’,
other 10 letters can be arranged between two N,
in which ‘O’ and ‘I’ repeat twice each and ‘T’ repeats 3 times.
∴ Required number of arrangements = \(\frac{10 !}{2 ! 2 ! 3 !}\)

Question 10.
Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements do not have the two R’s and two A’s together?
Solution:
There are 7 letters in the word ARRANGE in which ‘A’ and ‘R’ repeat 2 times each.
∴ Number of ways to arrange the letters of word ARRANGE = \(\frac{7 !}{2 ! 2 !}\) = 1260
Consider the words in which 2A are together and 2R are together.
Let us consider 2A as one unit and 2R as one unit.
These two units with remaining 3 letters can be arranged in = \(\frac{5 !}{2 ! 2 !}\) = 30 ways.
Number of arrangements in which neither 2A together nor 2R are together = 1260 – 30 = 1230

Question 11.
How many distinct 5 digit numbers can be formed using the digits 3, 2, 3, 2, 4, 5.
Solution:
5 digit numbers are to be formed from 2, 3, 2, 3, 4, 5.
Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5
Number of such numbers = \(\frac{5 !}{2 !}+\frac{5 !}{2 !}\) = 5! = 120
Case II: Numbers are formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers = \(\frac{5 !}{2 ! 2 !}+\frac{5 !}{2 ! 2 !}\) = 60
Required number of numbers = 120 + 60 = 180

Question 12.
Find the number of distinct numbers formed using the digits 3, 4, 5, 6, 7, 8, 9, so that odd positions are occupied by odd digits.
Solution:
A number is to be formed with digits 3, 4, 5, 6, 7, 8, 9 such that odd digits always occupy the odd places.
There are 4 odd digits, i.e. 3, 5, 7, 9.
∴ They can be arranged at 4 odd places among themselves in 4! = 24 ways.
There are 3 even digits, i.e. 4, 6, 8.
∴ They can be arranged at 3 even places among themselves in 3! = 6 ways.
∴ Required number of numbers formed = 24 × 6 = 144

Question 13.
How many different 6-digit numbers can be formed using digits in the number 659942? How many of them are divisible by 4?
Solution:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
∴ Required number of numbers formed = \(\frac{6 !}{2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360
A 6-digit number is to be formed using the same digits that are divisible by 4.
For a number to be divisible by 4, the last two digits should be divisible by 4,
i.e. 24, 52, 56, 64, 92 or 96.
Case I: When the last two digits are 24, 52, 56 or 64.
As the digit 9 repeats twice in the remaining four numbers, the number of arrangements = \(\frac{4 !}{2 !}\) = 12
∴ 6-digit numbers that are divisible by 4 so formed are 12 + 12 + 12 + 12 = 48.
Case II: When the last two digits are 92 or 96.
As each of the remaining four numbers are distinct, the number of arrangements = 4! = 24
∴ 6-digit numbers that are divisible by 4 so formed are 24 + 24 = 48.
∴ Required number of numbers framed = 48 + 48 = 96

Question 14.
Find the number of distinct words formed from letters in the word INDIAN. How many of them have the two N’s together?
Solution:
There are 6 letters in the word INDIAN in which I and N are repeated twice.
Number of different words that can be formed using the letters of the word INDIAN = \(\frac{6 !}{2 ! 2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 \times 2 !}\)
= 180
When two N’s are together.
Let us consider the two N’s as one unit.
They can be arranged with 4 other letters in \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60 ways.
∴ 2 N can be arranged in \(\frac{2 !}{2 !}\) = 1 way.
∴ Required number of words = 60 × 1 = 60

Question 15.
Find the number of different ways of arranging letters in the word PLATOON if (a) the two O’s are never together. (b) consonants and vowels occupy alternate positions.
Solution:
There are 7 letters in the words PLATOON in which ‘O’ repeat 2 times.
(a) When the two O’s are never together.
Let us arrange the other 5 letters first, which can be done in 5! = 120 ways.
The letters P, L, A, T, N create 6 gaps, in which O’s are arranged.
Two O’s can take their places in ⁶P₂ ways.
But ‘O’ repeats 2 times.
∴ Two O’s can be arranged in \(\frac{{ }^{6} \mathrm{P}_{2}}{2 !}\)
= \(\frac{\frac{6 !}{(6-2) !}}{2 !}\)
= \(\frac{6 \times 5 \times 4 !}{4 ! \times 2 \times 1}\)
= 3 × 5
= 15 ways
∴ Required number of arrangements = 120 × 15 = 1800

(b) When consonants and vowels occupy alternate positions.
There are 4 consonants and 3 vowels in the word PLATOON.
∴ At odd places, consonants occur and at even places, vowels occur.
4 consonants can be arranged among themselves in 4! ways.
3 vowels in which O occurs twice and A occurs once.
∴ They can be arranged in \(\frac{3 !}{2 !}\) ways.
Now, vowels and consonants should occupy alternate positions.
∴ Required number of arrangements = 4! × \(\frac{3 !}{2 !}\)
= 4 × 3 × 2 × \(\frac{3 \times 2 !}{2 !}\)
= 72

Class 11 Maharashtra State Board Maths Solution 

• Exercise 3.1 Class 11 Maths 2
• Exercise 3.2 Class 11 Maths 2
• Exercise 3.3 Class 11 Maths 2
• Exercise 3.4 Class 11 Maths 2
• Exercise 3.5 Class 11 Maths 2
• Exercise 3.6 Class 11 Maths 2
• Miscellaneous Exercise 3 Class 11 Maths 2

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.3 Questions and Answers.

11th Maths Part 2 Permutations and Combination Exercise 3.3 Questions And Answers Maharashtra Board

Question 1.
Find n, if ⁿP₆ : ⁿP₃ = 120 : 1.
Solution:
ⁿP₆ : ⁿP₃ = 120 : 1

∴ (n – 3) (n – 4) (n – 5) = 120
∴ (n – 3) (n – 4) (n – 5) = 6 × 5 × 4
Comparing on both sides, we get
n – 3 = 6
∴ n = 9

Question 2.
Find m and n, if ^(m+n)P₂ = 56 and ^(m-n)P₂ = 12.
Solution:
^(m+n)P₂ = 56

(m + n) (m + n – 1) = 56
Let m + n = t
t(t – 1) = 56
t² – t – 56 = 0
(t – 8) (t + 7) = 0
t = 8 or t = -7
m + n = 8 or m + n = -7
But m + n ≠ -7
∴ m + n = 8 ……(i)
Also, ^(m-n)P₂ = 12

(m – n) (m – n – 1) = 12
Let m – n = a
a(a – 1) = 12
a² – a – 12 = 0
(a – 4)(a + 3) = 0
a = 4 or a = -3
m – n = 4 or m – n = -3
But m – n ≠ -3
∴ m – n = 4 ……(ii)
Adding (i) and (ii), we get
2m = 12
∴ m = 6
Substituting m = 6 in (ii), we get
6 – n = 4
∴ n = 2

Question 3.
Find r, if ¹²P_r-2 : ¹¹P_r-1 = 3 : 14.
Solution:

(14 – r)(13 – r) = 8 × 7
Comparing on both sides, we get
14 – r = 8
∴ r = 6

Question 4.
Show that (n + 1) (ⁿP_r) = (n – r + 1) [⁽ⁿ⁺¹⁾P_r]
Solution:

Question 5.
How many 4 letter words can be formed using letters in the word MADHURI, if (a) letters can be repeated (b) letters cannot be repeated.
Solution:
There are 7 letters in the word MADHURI.
(a) A 4 letter word is to be formed from the letters of the word MADHURI and repetition of letters is allowed.
∴ 1st letter can be filled in 7 ways.
2nd letter can be filled in 7 ways.
3rd letter can be filled in 7 ways.
4th letter can be filled in 7 ways.
∴ Total no. of ways a 4-letter word can be formed = 7 × 7 × 7 × 7 = 2401

(b) When repetition of letters is not allowed, the number of 4-letter words formed from the letters of the word MADHURI is ⁷P₄ = \(\frac{7 !}{(7-4) !}=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}\) = 840

Question 6.
Determine the number of arrangements of letters of the word ALGORITHM if
(a) vowels are always together.
(b) no two vowels are together.
(c) consonants are at even positions.
(d) O is the first and T is the last letter.
Solution:
There are 9 letters in the word ALGORITHM.
(a) When vowels are always together.
There are 3 vowels in the word ALGORITHM (i.e., A, I, O).
Let us consider these 3 vowels as one unit.
This unit with 6 other letters is to be arranged.
∴ The number of arrangement = ⁷P₇ = 7! = 5040
3 vowels can be arranged among themselves in ³P₃ = 3! = 6 ways.
∴ Required number of arrangements = 7! × 3!
= 5040 × 6
= 30240

(b) When no two vowels are together.
There are 6 consonants in the word ALGORITHM,
they can be arranged among themselves in ⁶P₆ = 6! = 720 ways.
Let consonants be denoted by C.
_C _C_ C _C_C_C
There are 7 places marked by ‘_’ in which 3 vowels can be arranged.
∴ Vowels can be arranged in ⁷P₃ = \(\frac{7 !}{(7-3) !}=\frac{7 \times 6 \times 5 \times 4 !}{4 !}\) = 210 ways.
∴ Required number of arrangements = 720 × 210 = 151200

(c) When consonants are at even positions.
There are 4 even places and 6 consonants in the word ALGORITHM.
∴ 6 consonants can be arranged at 4 even positions in 6P4 = \(\frac{6 !}{(6-4) !}=\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\) = 360 ways.
Remaining 5 letters (3 vowels and 2 consonants) can be arranged in odd position in ⁵P₅ = 5! = 120 ways.
∴ Required number of arrangements = 360 × 120 = 43200

(d) When O is the first and T is the last letter.
All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T.
∴ Position of O and T are fixed.
∴ Other 7 letters can be arranged between O and T among themselves in ⁷P₇ = 7! = 5040 ways.
∴ Required number of arrangements = 5040

Question 7.
In a group photograph, 6 teachers and principal are in the first row and 18 students are in the second row. There are 12 boys and 6 girls among the students. If the middle position is reserved for the principal and if no two girls are together, find the number of arrangements.
Solution:
In 1st row, 6 teachers can be arranged among themselves in ⁶P₆ = 6! ways.
In the 2nd row, 12 boys can be arranged among themselves in ¹²P₁₂ = 12! ways.
No two girls are together.
So, there are 13 places formed by 12 boys in which 6 girls occupy any 6 places in ¹³P₆ ways.
∴ Required number of arrangements

Question 8.
Find the number of ways so that letters of the word HISTORY can be arranged as
(a) Y and T are together
(b) Y is next to T
(c) there is no restriction
(d) begin and end with a vowel
(e) end in ST
(f) begin with S and end with T
Solution:
There are 7 letters in the word HISTORY
(a) When ‘Y’ and ‘T’ are together.
Let us consider ‘Y’ and ‘T’ as one unit.
This unit with other 5 letters are to be arranged.
∴ The number of arrangement of one unit and 5 letters = ⁶P₆ = 6! = 720
Also, ‘Y’ and ‘T’ can be arranged among themselves in ²P₂ = 2! = 2 ways.
∴ A total number of arrangements when Y and T are always together = 6! × 2!
= 120 × 2
= 1440

(b) When ‘Y’ is next to ‘T’.
Let us take this (‘Y’ next to ‘T’) as one unit.
This unit with 5 other letters is to be arranged.
∴ The number of arrangements of 5 letters and one unit = ⁶P₆ = 6! = 720
Also, ‘Y’ has to be always next to ‘T’.
∴ They can be arranged among themselves in 1 way only.
∴ Total number of arrangements possible when Y is next to T = 720 × 1 = 720

(c) When there is no restriction.
7 letters can be arranged among themselves in ⁷P₇ = 7! ways.
∴ The total number of arrangements possible if there is no restriction = 7!

(d) When begin and end with a vowel.
There are 2 vowels in the word HISTORY.
All other letters of the word HISTORY are to be arranged between 2 vowels such that the arrangement begins and ends with a vowel.
The other 5 letters can be filled between the two vowels in ⁵P₅ = 5! = 120 ways.
Also, 2 vowels can be arranged among themselves at first and last places in ²P₂ = 2! = 2 ways.
∴ Total number of arrangements when the word begins and ends with vowel = 120 × 2 = 240

(e) When a word ends in ST.
As the arrangement ends with ST,
the remaining 5 letters can be arranged among themselves in ⁵P₅ = 5! = 120 ways.
∴ Total number of arrangements when the word ends with ST = 120

(f) When a word begins with S and ends with T.
As arrangement begins with S and ends with T,
the remaining 5 letters can be arranged between S and T among themselves in ⁵P₅ = 5! = 120 ways.
Total number of arrangements when the word begins with S and ends with T = 120

Question 9.
Find the number of arrangements of the letters in the word SOLAPUR so that consonants and vowels are placed alternately.
Solution:
There are 4 consonants S, L, P, R, and 3 vowels A, O, U in the word SOLAPUR.
Consonants and vowels are to be alternated.
∴ Vowels must occur in even places and consonants in odd places.
∴ 3 vowels can be arranged at 3 even places in ³P₃ = 3! = 6 ways.
Also, 4 consonants can be arranged at 4 odd places in ⁴P₄ = 4! = 24 ways.
Required number of arrangements = 6 × 24 = 144

Question 10.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 4, 5, 6, 8 if
(a) digits can be repeated.
(b) digits cannot be repeated.
Solution:
(a) A 4 digit number is to be made from the digits 1, 2, 4, 5, 6, 8 such that digits can be repeated.
∴ Unit’s place digit can be filled in 6 ways.
10’s place digit can be filled in 6 ways.
100’s place digit can be filled in 6 ways.
1000’s place digit can be filled in 6 ways.
∴ Total number of numbers that can be formed = 6 × 6 × 6 × 6 = 1296

(b) A 4 different digit number is to be made from the digits 1, 2, 4, 5, 6, 8 without repetition of digits.
∴ 4 different digits are to be arranged from 6 given digits which can be done in ⁶P₄ ways.
∴ Total number of numbers that can be formed
= \(\frac{6 !}{(6-4) !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360

Question 11.
How many numbers can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition so that resulting numbers are between 100 and 1000?
Solution:
A number between 100 and 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5 is of 3 digits, and repetition of digits is not allowed.
∴ 100’s place can be filled in 5 ways as it is a non-zero number.
10’s place digits can be filled in 5 ways.
Unit’s place digit can be filled in 4 ways.
∴ Total number of ways the number can be formed = 5 × 5 × 4 = 100

Question 12.
Find the number of 6-digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition. How many of these numbers are (a) divisible by 5 (b) not divisible by 5
Solution:
A number of 6 different digits is to be formed from the digits 3, 4, 5, 6, 7, 8 which can be done in ⁶P₆ = 6! = 720 ways.
(a) If the number is to be divisible by 5,
the unit’s place digit can be 5 only.
∴ it can be arranged in 1 way only.
The other 5 digits can be arranged among themselves in ⁵P₅ = 5! = 120 ways.
∴ Required number of numbers divisible by 5 = 1 × 120 = 120

(b) If the number is not divisible by 5,
unit’s place can be any digit from 3, 4, 6, 7, 8.
∴ it can be arranged in 5 ways.
Other 5 digits can be arranged in ⁵P₅ = 5! = 120 ways.
∴ Required number of numbers not divisible by 5 = 5 × 120 = 600

Question 13.
A code word is formed by two different English letters followed by two non-zero distinct digits. Find the number of such code words. Also, find the number of such code words that end with an even digit.
Solution:
There is a total of 26 alphabets.
A code word contains 2 English alphabets.
∴ 2 alphabets can be filled in ²⁶P₂
= \(\frac{26 !}{(26-2) !}\)
= \(\frac{26 \times 25 \times 24 !}{24 !}\)
= 650 ways.
Also, alphabets to be followed by two distinct non-zero digits from 1 to 9 which can be filled in
⁹P₂ = \(\frac{9 !}{(9-2) !}=\frac{9 \times 8 \times 7 !}{7 !}\) = 72 ways.
∴ Total number of a code words = 650 × 72 = 46800.
To find the number of codewords end with an even integer.
2 alphabets can be filled in 650 ways.
The digit in the unit’s place should be an even number between 1 to 9, which can be filled in 4 ways.
Also, 10’s place can be filled in 8 ways.
∴ Total number of codewords = 650 × 4 × 8 = 20800

Question 14.
Find the number of ways in which 5 letters can be posted in 3 post boxes if any number of letters can be posted in a post box.
Solution:
There are 5 letters and 3 post boxes and any number of letters can be posted in all three post boxes.
∴ Each letter can be posted in 3 ways.
∴ Total number of ways 5 letters can be posted = 3 × 3 × 3 × 3 × 3 = 243

Question 15.
Find the number of arranging 11 distinct objects taken 4 at a time so that a specified object (a) always occurs (b) never occurs
Solution:
There are 11 distinct objects and 4 are to be taken at a time.
(a) The number of permutations of n distinct objects, taken r at a time, when one particular object will always occur is \(\mathbf{r} \times{ }^{(\mathbf{n}-1)} \mathbf{P}_{(\mathbf{r}-1)}\)

∴ In 2880 permutations of 11 distinct objects, taken 4 at a time, one particular object will always occur.

(b) When one particular object will not occur, then 4 objects are to be arranged from 10 objects which can be done in ¹⁰P₄ = 10 × 9 × 8 × 7 = 5040 ways.
∴ In 5040 permutations of 11 distinct objects, taken 4 at a time, one particular object will never occur.

Question 16.
In how many ways can 5 different books be arranged on a shelf if
(i) there are no restrictions
(ii) 2 books are always together
(iii) 2 books are never together
Solution:
(i) 5 books arranged in ⁵P₅ = 5! = 120 ways.

(ii) 2 books are together.
Let us consider two books as one unit. This unit with the other 3 books can be arranged in ⁴P₄ = 4! = 24 ways.
Also, two books can be arranged among themselves in ²P₂ = 2 ways.
∴ Required number of arrangements = 24 × 2 = 48

(iii) Say books are B₁, B₂, B₃, B₄, B₅ are to be arranged with B₁, B₂ never together.
B₃, B₄, B₅ can be arranged among themselves in ³P₃ = 3! = 6 ways.
B₃, B₄, B₅ create 4 gaps in which B₁, B₂ are arranged in ⁴P₂ = 4 × 3 = 12 ways.
∴ Required number of arrangements = 6 × 12 = 72

Question 17.
3 boys and 3 girls are to sit in a row. How many ways can this be done if
(i) there are no restrictions.
(ii) there is a girl at each end.
(iii) boys and girls are at alternate places.
(iv) all-boys sit together.
Solution:
3 boys and 3 girls are to be arranged in a row.
(i) When there are no restrictions.
∴ Required number of arrangements = 6! = 720

(ii) When there is a girl at each end.
3 girls can be arranged at two ends in
³P₂ = \(\frac{3 !}{1 !}\) = 3 × 2 = 6 ways.
And remaining 1 girl and 3 boys can be arranged between the two girls in ⁴P₄ = 4! = 24 ways.
∴ Required number of arrangements = 6 × 24 = 144

(iii) Boys and girls are at alternate places.
We can first arrange 3 girls among themselves in ³P₃ = 3! = 6 ways.
Let girls be denoted by G.
G – G – G –
There are 3 places marked by ‘-’ where 3 boys can be arranged in 3! = 6 ways.
∴ Total number of such arrangements = 6 × 6 = 36
OR
Similarly, we can first arrange 3 boys in 3! = 6 ways
and then arrange 3 girls alternately in 3! = 6 ways.
∴ Total number of such arrangements = 6 × 6 = 36
∴ Required number of arrangements = 36 + 36 = 72

(iv) All boys sit together.
Let us consider all boys as one group.
This one group with the other 3 girls can be arranged ⁴P₄ = 4! = 24 ways.
Also, 3 boys can be arranged among themselves in ³P₃ = 3! = 6 ways.
∴ Required number of arrangements = 24 × 6 = 144

Class 11 Maharashtra State Board Maths Solution 

• Exercise 3.1 Class 11 Maths 2
• Exercise 3.2 Class 11 Maths 2
• Exercise 3.3 Class 11 Maths 2
• Exercise 3.4 Class 11 Maths 2
• Exercise 3.5 Class 11 Maths 2
• Exercise 3.6 Class 11 Maths 2
• Miscellaneous Exercise 3 Class 11 Maths 2