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Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.4

Question 1.
Find the number of permutations of letters in each of the following words:
(i) DIVYA
(ii) SHANTARAM
(iii) REPRESENT
(iv) COMBINE
Solution:
(i) There are 5 letters in the word DIVYA which can be arranged in 5! Way = 120 ways

(ii) There are 9 letters in the word SHANTARAM in which ‘A’ repeats 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’ repeats 3 times and ‘R’ repeats 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 which can be arranged among themselves in 7! = 5040 ways

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 = \(\frac{9 !}{2 ! 3 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 3 \times 2 \times 4 !}\)
= 9 × 4 × 7 × 5
= 1260
∴ In 1260 distinct ways the books can be arranged on a shelf.

Question 3.
A coin is tossed 8 times. In how many ways can we obtain
(i) 4 heads and 4 tails?
(ii) at least 6 heads?
Solution:
A coin is tossed 8 times. All heads are identical and all tails are identical.
(i) We can obtain 4 heads and 4 tails in \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2}\)
= 70 ways
∴ In 70 different ways we can obtain 4 heads and 4 tails.

(ii) When 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 in which 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
∴ In 37 different ways we can obtain at least 6 heads.

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’ repeats 2 times, ‘A’ repeats 3 times, and ‘T’ repeats 2 times.
∴ Total number of arrangements = \(\frac{12 !}{2 ! 3 ! 2 !}\)
When all the vowels
i.e., ‘A’, ‘A’, ‘A’, ‘E’, T are to be kept together
Number of arrangements of these vowels = \(\frac{5 !}{3 !}\) ways.
Let us consider these vowels together as one unit.
This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each.
∴ Number of arrangements = \(\frac{8 !}{2 ! 2 !}\)
∴ Total 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
(i) letters M and T never together?
(ii) 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 arrangements is \(\frac{11 !}{4 ! 2 ! 2 !}\)
∴ \(\frac{11 !}{4 ! 2 ! 2 !}\) different words can be formed from the letters of the word MAHARASHTRA.
(i) Other than M and T. there are 9 letters in which A repeats 4 times, H repeats twice, R repeats twice
The number of arrangements of the a letter = \(\frac{9 !}{4 ! 2 ! 2 !}\)
These 9 letters create 10 gaps in which M and T are to be arranged
The number of arrangements of M and T = ¹⁰P₂
∴ Total number arrangement having M and T never together = \(\frac{9 ! \times{ }^{10} \mathrm{P}_{2}}{4 ! 2 ! 2 !}\)

(ii) 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, they themselves can be arranged in \(\frac{4 !}{4 !}\) = 1 way.
This unit is to be arranged with 7 other letters which can be done in 8! ways
∴ Total number of arrangements = \(\frac{8 !}{2 ! 2 !}\)
∴ \(\frac{8 !}{2 ! 2 !}\) different words can be formed if vowels are always together.

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:
When ‘R’ is used thrice, ‘S’ is used twice and ‘T’ is used twice,
∴ Total number of letters available = 7, of which ‘S’ and ‘T’ repeat 2 times each, ‘R’ repeats 3 times.
∴ Required number of arrangements = \(\frac{7 !}{2 ! 2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 1 \times 2 \times 1 \times 3 !}\)
= 7 × 6 × 5
= 210
∴ 210 different words can be formed with the letter R is used thrice and letters S and T are used twice each.

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 other 4 letters in which ‘M’ repeats twice, is to be arranged.
∴ Total number of arrangements when B is always next to A = \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 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’, T repeat two times each, ‘T’ repeats 3 times.
The arrangement starts and ends with ‘N’, 10 letters other than N can be arranged between two N, in which ‘O’ and ‘I’ repeat twice each and ‘T’ repeats 3 times.
∴ Total number of arrangements with the letter N at the beginning and at the end = \(\frac{10 !}{2 ! 2 ! 3 !}\)

Question 10.
Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements have two R’s and two A’s not together?
Solution:
(i) There are 7 letters in the word ARRANGE in which A is repeated 2 times and R is repeated 2 times
∴ The number of arrangements = \(\frac{7 !}{2 ! 2 !}\) = 1260

(ii) A: set of words having 2A together
B: set of words having 2R together
Number of words having both A and both R not together
= 1260 – n(A ∪ B)
= 1260 – [n(a) + n(B) – n(A ∩ B)] ……(i)
n(A) = number of ways in which (AA) R, R, N, G, E are to be arranged
∴ n(A) = \(\frac{6 !}{2 !}\) = 360
n(B) = number of ways in which (RR), A, A, N, G, E are to be arranged
∴ n(B) = \(\frac{6 !}{2 !}\) = 360
n(A ∩ B) = number of ways in which (AA), (RR), N, G, E are to be arranged
∴ n(A ∩ B) = 5! = 120
Substituting n(A), n(B), n(A ∩ B) in (i), we get
Number of words having both A and both R not together
= 1260 – [360 + 360 – 120]
= 1260 – 600
= 660

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 = \(\frac{5 !}{2 !}\) × 2
= 5!
= 120

Case II: Numbers formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers = \(\frac{5 !}{2 ! 2 !}\) × 2 = 60
Required number = 120 + 60 = 180
∴ 180 distinct 5 digit numbers can be formed using the digit 3, 2, 3, 2, 4, 5.

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! ways = 24 ways
3 even places of the number are occupied by even digits (i.e. 4, 6, 8).
∴ They can be arranged in 3! ways = 6 ways
∴ Total number of arrangements = 24 × 6 = 144
∴ 144 numbers can be formed so that odd digits always occupy the odd positions.

Question 13.
How many different 6-digit numbers can be formed using digits in the number 659942? How many of them are divisible by 2?
Solution:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
∴ Total number of arrangements = \(\frac{6 !}{2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360
∴ 360 different 6-digit numbers can be formed.
For a number to be divisible by 2,
Last digits can be selected in 3 ways
Remaining 5 digit in which, 9 appears twice are arranged in \(\frac{5 !}{2 !}\) ways
∴ Total number of arrangements = \(\frac{5 !}{2 !}\) × 3 = 180
∴ 180 numbers are divisible by 2.

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 repeat 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
∴ 180 different words can be formed with the letters of the word INDIAN.
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.
∴ 2N can be arranged in 1 way
∴ Total number of arrangements = 60 × 1 = 60 ways
∴ 60 words are such that two N’s are together.

Question 15.
Find the number of different ways of arranging letters in the word PLATOON if
(i) the two O’s are never together.
(ii) consonants and vowels occupy alternate positions.
Solution:
(i) 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 in 6 gaps can be arranged in \(\frac{{ }^{6} \mathrm{P}_{2}}{2 !}\) ways
= \(\frac{\frac{6 !}{(6-2) !}}{2 !}\) ways
= \(\frac{6 \times 5 \times 4 !}{4 ! \times 2 \times 1}\) ways
= 3 × 5 ways
= 15 ways
∴ Total number of arrangements if the two O’s are never together = 120 × 15 = 1800

(ii) 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
∴ Required number of arrangements if the consonants and vowels occupy alternate positions = 4! × \(\frac{3 !}{2 !}\)
= 4 × 3 × 2 × \(\frac{3 \times 2 !}{2 !}\)
= 72

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.3

Question 1.
Find n if ⁿP₆ : ⁿP₃ = 120 : 1
Solution:
ⁿP₆ : ⁿP₃ = 120 : 1
∴ \(\frac{n !}{(n-6) !} \div \frac{n !}{(n-3) !}=\frac{120}{1}\)
∴ \(\frac{\mathrm{n} !}{(\mathrm{n}-6) !} \times \frac{(\mathrm{n}-3) !}{\mathrm{n} !}\) = 120
∴ \(\frac{n !}{(n-6) !} \times \frac{(n-3)(n-4)(n-5)(n-6) !}{n !}\) = 120
∴ (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+nP₂ = 56
∴ \(\frac{(\mathrm{m}+\mathrm{n}) !}{(\mathrm{m}+\mathrm{n}-2) !}\) = 56
∴ \(\frac{(m+n)(m+n-1)(m+n-2) !}{(m+n-2) !}\) = 56
∴ (m + n) (m + n – 1) = 8 × 7
Comparing on both sides, we get
m + n = 8 …..(i)
Also ^m-nP₂ = 12
∴ \(\frac{(m-n) !}{(m-n-2) !}\) = 12
∴ \(\frac{(m-n)(m-n-1)(m-n-2) !}{(m-n-2) !}\) = 12
∴ (m – n) (m – n – 1) = 4 × 3
Comparing on both sides, we get
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
(i) letters can be repeated?
(ii) letters cannot be repeated?
Solution:
There are 7 letters in the word MADHURI.
(i) 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
∴ 2401 four-lettered words can be formed when repetition of letters is allowed.

(ii) 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
∴ 840 four-letter words can be formed when repetition of letters is not allowed.

Alternate method:
There are 7 letters in the word MADHURI.
(i) Since letters can be repeated
∴ In all places of a four-letter word, any one of seven letters M, A, D, H, U, R, I can appear.
∴ Using the Multiplication theorem, we get
Number of four-letter words with repetition of letters M, A, D, H, U, R, I = 7 × 7 × 7 × 7 = 2401

(ii) Since the letters cannot be repeated therefore 1st, 2nd, 3rd, 4th places can be filled in 7, 6, 5, 4 ways respectively
∴ Using the multiplication theorem, we get
Number of four-letter words, with no repetition of letters M, A, D, H, U, R, I = 7 × 6 × 5 × 4 = 840

Question 6.
Determine the number of arrangements of letters of the word ALGORITHM if
(i) vowels are always together.
(ii) no two vowels are together.
(iii) Consonants are at even positions
(iv) O is first and T is last.
Solution:
A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.
(i) 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.
∴ It becomes an arrangement of 7 things which can be done in ⁷P₇ i.e., 7! ways and 3 vowels can be arranged among themselves in ³P₃ i.e., 3! ways.
∴ the total number of ways in which the word can be formed = 7! × 3!
= 5040 × 6
= 30240
∴ 30240 words can be formed if vowels are always together.

(ii) When no two vowels are together:
There are 6 consonants in the word ALGORITHM.
They can be arranged among themselves in ⁶P₆ i.e., 6! ways.
Let consonants be denoted by C.
_C_C_ C_C_C_C_
6 consonants create 7 gaps in which 3 vowels are to arranged.
∴ 3 vowels can be filled in ⁷P₃
= \(\frac{7 !}{(7-3) !}\)
= \(\frac{7 \times 6 \times 5 \times 4 !}{4 !}\)
= 210 ways
∴ total number of ways in which the word can be formed = 6! × 210
= 720 × 210
= 151200
∴ 151200 words can be formed if no two vowels are together.

(iii) When consonants are at even positions:
There are 4 even places and 6 consonants in the word ALGORITHM.
1st, 2nd, 3rd, 4th even places are filled in 6, 5, 4, 3 way respectively.
∴ The number of ways to fill four even places by consonants = 6 × 5 × 4 × 3 = 360
The remaining 5 letters (3 vowels and 2 consonants) can be arranged among themselves in ⁵P₅ i.e., 5! ways.
∴ Total number of ways the words can be formed
In which even places are occupied by consonants = 360 × 5!
= 360 × 120
= 43200
∴ 43200 words can be formed if even positions are occupied by consonants.

(iv) When beginning with O and ends with T:
All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T.
7 letters other than O and T can be filled between O and T in ⁷P₇ i.e., 7! ways = 5040 ways.
∴ 5040 words beginning with O and ending with T can be formed.

Question 7.
In a group photograph, 6 teachers and principals 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 middle seat is fixed for the principal.
Also 1st row, 6 teachers can be arranged among themselves in ⁶P₆ i.e., 6! ways.
In the 2nd row, 12 boys can be arranged among themselves in ¹²P₁₂ i.e., 12! ways.
13 gaps are created by 12 boys, in which 6 girls are to be arranged.
together which can be done in ¹³P₆ ways.
∴ total number of arrangements = 6! × 12! × ¹³P₆ …..[using Multiplications Principle]
= 6! × 12! × \(\frac{13 !}{(13-6) !}\)
= 6! × 12! × \(\frac{13 !}{7 !}\)
= \(\frac{6 ! \times 12 ! \times 13 !}{7 \times 6 !}\)
= \(\frac{12 ! 13 !}{7}\)

Question 8.
Find the number of ways letters of the word HISTORY can be arranged if
(i) Y and T are together
(ii) Y is next to T.
Solution:
There are 7 letters in the word HISTORY
(i) When ‘Y’ and ‘T’ are together.
Let us consider ‘ Y’ and ‘T’ as one unit
This unit with the other 5 letters is to be arranged.
∴ The number of arrangements of one unit and 5 letters = ⁶P₆ = 6!
Also, ‘Y’ and ‘T’ can be arranged among themselves in ²P₂ i.e., 2! ways.
∴ a total number of arrangements when Y and T are always together = 6! × 2!
= 720 × 2
= 1440
∴ 1440 words can be formed if Y and T are together.

(ii) 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 6 letters and one unit = ⁶P₆ = 6!
Also ‘Y’ has to be always next to ‘T’.
So they can be arranged in 1 way.
∴ total number of arrangements possible when Y is next to T = 6! × 1 = 720
∴ 720 words can be formed if Y is next to T.

Question 9.
Find the number of arrangements of the letters in the word BERMUDA so that consonants and vowels are in the same relative positions.
Solution:
There are 7 letters in the word “BERMUDA” out of which 3 are vowels and 4 are consonants.
If relative positions of consonants and vowels are not changed.
3 vowels can be arranged among themselves in ³P₃ i.e., 3! ways.
4 consonants can be arranged among themselves in ⁴P₄ i.e., 4! ways.
∴ total no. of arrangements possible if relative positions of vowels and consonants are not changed = 3! × 4!
= 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
(i) digits can be repeated
(ii) digits cannot be repeated
Solution:
(i) A 4 digit number is to be made from the digits 1, 2, 4, 5, 6, 8 such that digits can be repeated.
∴ The 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 = 6 × 6 × 6 × 6 = 6⁴ = 1296
∴ 1296 four-digit numbers can be formed if repetition of digits is allowed.

(ii) 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
= \(\frac{6 !}{(6-4) !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360 ways
∴ 360 four-digit numbers can be formed, if repetition of digits is not allowed.

Question 11.
How many numbers can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition so that the resulting numbers are between 100 and 1000?
Solution:
A number between 100 and 1000 is a 3 digit number and is to be formed from the digits 0, 1, 2, 3, 4, 5, without repetition of digits.
∴ 100’s place digit must be a non-zero number which can be filled in 5 ways.
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
∴ 100 numbers between 100 and 1000 can be formed.

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
(i) divisible by 5?
(ii) 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 i.e., 6! = 720 ways
(i) If the number is divisible by 5, then
The unit’s place digit must be 5, and hence unit’s place can be filled in 1 way
Other 5 digits can be arranged among themselves in ⁵P₅ i.e., 5! ways
∴ Total number of ways in which numbers divisible by 5 can be formed = 1 × 5! = 120

(ii) If the number is not divisible by 5, then
Unit’s place can be any digit from 3, 4, 6, 7, 8 which can be selected in 5 ways.
Other 5 digits can be arranged in ⁵P₅ i.e., 5! ways
∴ The total number of ways in which numbers not divisible by 5 can be formed = 5 × 5!
= 5 × 120
= 600

Question 13.
A code word is formed by two distinct 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:
(i) 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

(ii) There are total 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
For a code word to end with an even integer, 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 ways
∴ 20800 codewords end with an even integer.

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 in which 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
(i) always occurs
(ii) never occurs
Solution:
There are 11 distinct objects and 4 are to be taken at a time.
(i) The number of permutations of n distinct objects, taken r at a time, when one specified object will always occur is r × ^(n-1)P_(r-1)
Here, r = 4, n = 11
∴ The number of permutations of 4 out of 11 objects when a specified object occurs.
= 4 × ^(11-1)P_(4-1)
= 4 × ¹⁰P₃
= 4 × \(\frac{10 !}{(10-3) !}\)
= 4 × \(\frac{10 !}{7 !}\)
= 4 × \(\frac{10 \times 9 \times 8 \times 7 !}{7 !}\)
= 2880
∴ There are 2880 permutations of 11 distinct objects, taken 4 at a time, in which one specified object always occurs.

(ii) When one specified object does not occur then 4 things are to be arranged from the remaining 10 things, which can be done in ¹⁰P₄ ways
= 10 × 9 × 8 × 7 ways
= 5040 ways
∴ There are 5040 permutations of 11 distinct objects, taken 4 at a time, in which one specified object never occurs.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.2

Question 1.
Evaluate:
(i) 8!
Solution:
8!
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320

(ii) 6!
Solution:
6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

(iii) 8! – 6!
Solution:
8! – 6!
= 8 × 7 × 6! – 6!
= 6! (8 × 7 – 1)
= 6! (56 – 1)
= 6 × 5 × 4 × 3 × 2 × 1 × 55
= 39,600

(iv) (8 – 6)!
Solution:
(8 – 6)!
= 2!
= 2 × 1
= 2

Question 2.
Compute:
(i) \(\frac{12 !}{6 !}\)
Solution:
\(\frac{12 !}{6 !}\)
= \(\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\)
= 12 × 11 × 10 × 9 × 8 × 7
= 665280

(ii) \(\left(\frac{12}{6}\right) !\)
Solution:
\(\left(\frac{12}{6}\right) !\)
= 2!
= 2 × 1
= 2

(iii) (3 × 2)!
Solution:
(3 × 2)!
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

(iv) 3! × 2!
Solution:
3! × 2!
= 3 × 2 × 1 × 2 × 1
= 12

Question 3.
Compute:
(i) \(\frac{9 !}{3 ! 6 !}\)
Solution:

(ii) \(\frac{6 !-4 !}{4 !}\)
Solution:

(iii) \(\frac{8 !}{6 !-4 !}\)
Solution:

(iv) \(\frac{8 !}{(6-4) !}\)
Solution:

Question 4.
Write in terms of factorials
(i) 5 × 6 × 7 × 8 × 9 × 10
Solution:
5 × 6 × 7 × 8 × 9 × 10
= 10 × 9 × 8 × 7 × 6 × 5
Multiplying and dividing by 4!, we get

(ii) 3 × 6 × 9 × 12 × 15
Solution:
3 × 6 × 9 × 12 × 15
= 3 × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5)
= (3⁵) (5 × 4 × 3 × 2 × 1)
= 3⁵ (5!)

(iii) 6 × 7 × 8 × 9
Solution:
6 × 7 × 8 × 9
= 9 × 8 × 7 × 6
Multiplying and dividing by 5!, we get

(iv) 5 × 10 × 15 × 20 × 25
Solution:
5 × 10 × 15 × 20 × 25
= (5 × 1) × (5 × 2) × (5 × 3) × (5 × 4) × (5 × 5)
= (5⁵) (5 × 4 × 3 × 2 × 1)
= (5⁵) (5!)

Question 5.
Evaluate: \(\frac{n !}{r !(n-r) !}\) for
(i) n = 8, r = 6
Solution:
n = 8, r = 6

(ii) n = 12, r = 12
Solution:
n = 12, r = 12

Question 6.
Find n, if
(i) \(\frac{n}{8 !}=\frac{3}{6 !}+\frac{1}{4 !}\)
Solution:

(ii) \(\frac{n}{6 !}=\frac{4}{8 !}+\frac{3}{6 !}\)
Solution:

(iii) \(\frac{1}{n !}=\frac{1}{4 !}-\frac{4}{5 !}\)
Solution:

Question 7.
Find n, if
(i) (n + 1)! = 42 × (n – 1)!
Solution:
(n + 1)! = 42(n – 1)!
∴ (n + 1) n (n – 1)! = 42(n – 1)!
∴ n² + n = 42
∴ n(n + 1) = 6 × 7
Comparing on both sides, we get
∴ n = 6

(ii) (n + 3)! = 110 × (n + 1)!
Solution:
(n + 3)! = 110 × (n + 1)!
∴ (n + 3) (n + 2) (n + 1)! = 110 (n + 1)!
∴ (n + 3) (n + 2) = (11) (10)
Comparing on both sides, we get
n + 3 = 11
∴ n = 8

Question 8.
Find n, if:
(i) \(\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-5) !}=5: 3\)
Solution:

∴ 12 = (n – 3)(n – 4)
∴ (n – 3)(n – 4) = 4 × 3
Comparing on both sides, we get
n – 3 = 4
∴ n = 7

(ii) \(\frac{n !}{3 !(n-5) !}: \frac{n !}{5 !(n-7) !}=10: 3\)
Solution:

∴ (n – 5) (n – 6) = 3 × 2
Comparing on both sides, we get
n – 5 = 3
∴ n = 8

Question 9.
Find n, if:
(i) \(\frac{(17-n) !}{(14-n) !}\) = 5!
Solution:
\(\frac{(17-n) !}{(14-n) !}\) = 5!
∴ \(\frac{(17-n)(16-n)(15-n)(14-n) !}{(14-n) !}\) = 5 × 4 × 3 × 2 × 1
∴ (17 – n) (16 – n) (15 – n) = 6 × 5 × 4
Comparing on both sides, we get
17 – n = 6
∴ n = 11

(ii) \(\frac{(15-n) !}{(13-n) !}\) = 12
Solution:
\(\frac{(15-n) !}{(13-n) !}\) = 12
∴ \(\frac{(15-\mathrm{n})(14-\mathrm{n})(13-\mathrm{n}) !}{(13-\mathrm{n}) !}\) = 12
∴ (15 – n) (14 – n) = 4 × 3
Comparing on both sides, we get
15 – n = 4
∴ n = 11

Question 10.
Find n if \(\frac{(2 n) !}{7 !(2 n-7) !}: \frac{n !}{4 !(n-4) !}\) = 24 : 1
Solution:

∴ (2n – 1) (2n – 3) (2n – 5) = 9 × 7 × 5
Comparing on both sides, we get
2n – 1 = 9
∴ n = 5

Question 11.
Show that \(\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !}=\frac{(n+1) !}{r !(n-r+1)}\)
Solution:

Question 12.
Show that \(\frac{9 !}{3 ! 6 !}+\frac{9 !}{4 ! 5 !}=\frac{10 !}{4 ! 6 !}\)
Solution:

Question 13.
Find the value of:
(i) \(\frac{8 !+5(4 !)}{4 !-12}\)
Solution:

(ii) \(\frac{5(26 !)+(27 !)}{4(27 !)-8(26 !)}\)
Solution:

Question 14.
Show that
\(\frac{(2 n) !}{n !}\) = 2ⁿ (2n – 1) (2n – 3)…5.3.1
Solution:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.1

Question 1.
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?
Solution:
There are 30 boys and 20 girls in a class.
The teacher wants to select a class monitor from these boys and girls.
A boy can be selected in 30 ways and a girl can be selected in 20 ways.
∴ By using the fundamental principle of addition,
in a number of ways either a boy or a girl is selected as a class monitor = 30 + 20 = 50.

Question 2.
In question 1, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?
Solution:
(i) Since there are 30 boys in the class
∴ A boy monitor can be selected in 30 ways.
(ii) Since there are 20 girls in the class
∴ A girl monitor can be selected in 20 ways.

Question 3.
A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?
Solution:
A signal is generated from 2 flags and there are 4 flags of different colours available.
∴ 1st flag can be any one of the available 4 flags.
∴ It can be selected in 4 ways.
Now, 2nd flag is to be selected for which 3 flags are available for a different signal.
∴ 2nd flag can be anyone from these 3 flags.
∴ It can be selected in 3 ways.
∴ By using the fundamental principle of multiplication,
Total number of ways in which a signal can be generated = 4 × 3 = 12
∴ 12 different signals can be generated.

Question 4.
How many two-letter words can be formed using letters from the word SPACE when repetition of letters
(i) is allowed
(ii) is not allowed
Solution:
A two-letter word is to be formed out of the letters of the word SPACE.
(i) When repetition of the letters is allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 5 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 5 = 25

(ii) When repetition of the letters is not allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 4 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 4 = 20

Question 5.
How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits
(i) are allowed
(ii) are not allowed
Solution:
The three-digit number is to be formed from the digits 0, 1, 3, 5, 6
(i) When repetition of digits is allowed:
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6
∴ 100’s place digit can be selected in 4 ways.
0 can appear in 10’s and unit’s place and digits can be repeated.
∴ 10’s place digit can be selected in 5 ways and the unit’s place digit can be selected in 5 ways.
∴ By using the fundamental principle of multiplication,
the total number of three-digit numbers = 4 × 5 × 5 = 100

(ii) When repetition of digits is not allowed:
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6
∴ 100’s place digit can be selected in 4 ways
0 can appear in 10’s and unit’s place and digits can’t be repeated.
∴ 10’s place digit can be selected in 4 ways and the unit’s place digit can be selected in 3 ways
∴ By using the fundamental principle of multiplication,
total number of three-digit numbers = 4 × 4 × 3 = 48

Question 6.
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Solution:
A 3-digit number is to be formed from the digits 2, 3, 4, 5, 6 where digits can be repeated.
∴ The unit’s place digit can be selected in 5 ways.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 5 ways.
∴ By using fundamental principle of multiplication,
the total number of 3-digit numbers = 5 × 5 × 5 = 125

Question 7.
A letter lock has 3 rings and each ring has 5 letters. Determine the maximum number of trials that may be required to open the lock.
Solution:
A letter lock has 3 rings, each ring containing 5 different letters.
∴ A letter from each ring can be selected in 5 ways.
∴ By using fundamental principle of multiplication,
the total number of trials that can be made = 5 × 5 × 5 = 125
Out of these 124 wrong attempts are made and in the 125th attempt,
the lock gets opened, for a maximum number of trials.
∴ A maximum number of trials required to open the lock is 125.

Question 8.
In a test that has 5 true/false questions, no student has got all correct answers and no sequence of answers is repeated. What is the maximum number of students for this to be possible?
Solution:
For a set of 5 true/false questions, each question can be answered in 2 ways.
∴ By using fundamental principle of multiplication,
the total number of possible sequences of answers = 2 × 2 × 2 × 2 × 2 = 32
Since no student has written all the correct answers.
∴ Total number of sequences of answers given by the students in the class = 32 – 1 = 31
Also, no student has given the same sequence of answers.
∴ Maximum number of students in the class = Number of sequences of answers given by the students = 31

Question 9.
How many numbers between 100 and 1000 have 4 in the unit’s place?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 where the unit place digit is 4.
Since Unit’s place digit is 4.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 10 ways.
For 3-digit number 100’s place digit should be a non-zero number.
∴ 100’s place digit can be selected in 9 ways.
∴ By using fundamental principle of multiplication,
total number of numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90

Question 10.
How many numbers between 100 and 1000 have the digit 7 exactly once?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.
When 7 is in the unit’s place
The unit’s place digit is 7.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 9 ways.
100’s place digit can be selected in 8 ways.
∴ total number of numbers which have 7 in the unit’s place = 1 × 9 × 8 = 72
When 7 is in 10’s place
The unit’s place digit can be selected in 9 ways.
10’s place digit is 7
∴ it can be selected in 1 way only.
100’s place digit can be selected in 8 ways.
∴ total number of numbers which have 7 in 10’s place = 9 × 1 × 8 = 72
When 7 is in 100’s place
The unit’s place digit can be selected in 9 ways.
10’s place digit can be selected in 9 ways.
100’s place digit is 7
∴ it can be selected in 1 way.
∴ total numbers which have 7 in 100’s place = 9 × 9 × 1 = 81
∴ total number of numbers between 100 and 1000 having digit 7 exactly once = 72 + 72 + 81 = 225.

Question 11.
How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?
Solution:
Among many set’s of digits, the greatest number is possible when digits are arranged in descending order.
∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.
∴ Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed.
∴ 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Total number of numbers not exceeding 7432 that can be formed from the digits 2, 3, 4, 7
= Total number of four-digit numbers formed from the digits 2, 3, 4, 7
= 4 × 3 × 2 × 1
= 24

Question 12.
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Solution:
Case I: Three-digit numbers with 4 occurring in hundred’s place:
100’s place digit can be selected in 1 way.
Ten’s place can be filled by any one of the numbers 2, 3, 5, 6.
∴ 10’s place digit can be selected in 4 ways.
The unit’s place digit can be selected in 3 ways.
∴ total number of numbers which have 4 in 100’s place = 1 × 4 × 3 = 12

Case II: Three-digit numbers more than 500
100’s place digit can be selected in 2 ways.
10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ total number of three digit numbers more than 500 = 2 × 4 × 3 = 24

Case III: Number of four digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ total four digit numbers formed = 5 × 4 × 3 × 2 = 120

Case IV: Number of five digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ total five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120
∴ total number of numbers that exceed 400 = 12 + 24 + 120 + 120 = 276

Question 13.
How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?
Solution:
Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8
Number of such numbers = 3

Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8
Ten’s place digit is selected from 2, 5, 7, 8.
∴ Ten’s place digit can be selected in 4 ways.
Unit’s place digit is anyone from 0, 1, 2, 5, 7, 8
∴ The unit’s place digit can be selected in 6 ways.
Using the multiplication principle,
the number of such numbers (repetition allowed) = 4 × 6 = 24

Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8
100’s place digit is anyone from 1, 2, 5, 7, 8.
∴ 100’s place digit can be selected in 5 ways.
As digits can be repeated, the 10’s place and unit’s place digits are selected from 0, 1, 2, 5, 7, 8
∴ 10’s place and unit’s place digits can be selected in 6 ways each.
Using multiplication principle,
the number of such numbers (repetition allowed) = 5 × 6 × 6 = 180
All cases are mutually exclusive and exhaustive.
∴ Required number = 3 + 24 + 180 = 207

Question 14.
A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?
Solution:
A student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.
∴ Number of ways to choose gates = 3
Number of ways to choose staircase = 4
∴ By using fundamental principle of multiplication,
number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12

Question 15.
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
Solution:
For a number to be divisible by 3.
The sum of digits must be divisible by 3.
Given 6 digits are 0, 1,2, 3, 4, 5.
Sum of 1, 2, 3, 4, 5 = 15, which is divisible by 3.
∴ There are two cases of 5 digit numbers formed from 0, 1, 2, 3, 4, 5 and divisible by 3.
Either 3 is selected in 5 digits (and 0 not selected) or 3 is not selected in 5 digits (and 0 is selected)
Case I:
3 is not selected (and 0 is selected) i.e., the digits are 0, 1, 2, 4, 5.
10000’s place digit can be selected in 4 ways (as 0 cannot appear).
As digits are not repeated, 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Using multiplication theorem,
Number of 5-digit number formed from 0, 1, 2, 4, 5 (with no repetition of digits) = 4 × 4 × 3 × 2 × 1 = 96

Case II:
3 is selected (and 0 is not selected) i.e., 1, 2, 3, 4, 5
10000’s place digit can be selected in 5 ways.
1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
Using multiplication theorem,
Number of 5-digit numbers formed from 1, 2, 3, 4, 5 = 5 × 4 × 3 × 2 × 1 = 120
Both the cases are mutually exclusive and exhaustive.
∴ Required number = 96 + 120 = 216

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Correlation Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Correlation Miscellaneous Exercise 5

Question 1.
Two series of x and y with 50 items each have standard deviations of 4.8 and 3.5 respectively. If the sum of products of deviations of x and y series from respective arithmetic means is 420, then find the correlation coefficient between x and y.
Solution:

Question 2.
Find the number of pairs of observations from the following data,
r = 0.15, σ_y = 4, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 12, \(\Sigma\left(x_{i}-\bar{x}\right)^{2}\) = 40.
Solution:
Given, r = 0.15, σ_y = 4, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 12, \(\Sigma\left(x_{i}-\bar{x}\right)^{2}\) = 40

Question 3.
Given that r = 0.4, σ_y = 3, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 108, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)^{2}\) = 900. Find the number of pairs of observations.
Solution:
Given, r = 0.4, σ_y = 3, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 108, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)^{2}\) = 900

Question 4.
Given the following information, Σ\(x_{\mathrm{i}}^{2}\) = 90, Σx_iy_i = 60, r = 0.8, σ_y = 2.5, where x_i and y_i are the deviations from their respective means, find the number of items.
Solution:
Here, Σ\(x_{\mathrm{i}}^{2}\) = 90, Σx_iy_i = 60, r = 0.8, σ_y = 2.5
Here, x_i and y_i are the deviations from their respective means.
∴ If X_i, Y_i are elements of x and y series respectively, then
X_i – \(\bar{x}\) = x_i and Y_i – \(\bar{y}\) = y_i
∴ Σx_iy_i = \(\Sigma\left(\mathrm{X}_{\mathrm{i}}-\bar{x}\right)\left(\mathrm{Y}_{\mathrm{i}}-\bar{y}\right)\) = 60, \(\sum x_{\mathrm{i}}^{2}=\sum\left(\mathrm{X}_{\mathrm{i}}-\bar{x}\right)^{2}\) = 90

Question 5.
A sample of 5 items is taken from the production of a firm. The length and weight of 5 items are given below. [Given: √0.8823 = 0.9393]

Calculate the correlation coefficient between length and weight and interpret the result.
Solution:
Let length = x_i (in cm), Weight = y_i (in gm)


∴ the value of r indicates a high degree of positive correlation between length and weight of items.

Question 6.
Calculate the correlation coefficient from the following data and interpret it.

Solution:


∴ the value of r indicates a perfect negative correlation between x and y.

Question 7.
Calculate the correlation coefficient from the following data and interpret it.

Solution:



∴ the value of r indicates a perfect positive correlation between x and y.

Question 8.
If the correlation coefficient between X and Y is 0.8, what is the correlation coefficient between
(i) 2X and Y
(ii) \(\frac{X}{2}\) and Y
(iii) X and 3Y
(iv) X – 5 and Y – 3
(v) X + 7 and Y + 9
(vi) \(\frac{X-5}{7}\) and \(\frac{Y-3}{8}\)?
Solution:
The correlation coefficient remains unaffected by the change of origin and scale.
i.e., if u_i = \(\frac{x_{i}-\mathrm{a}}{\mathrm{h}}\) and v_i = \(\frac{y_{i}-\mathrm{b}}{\mathrm{k}}\), then Corr(U, V) = ±Corr(X, Y).
according to the same or opposite signs of h and k.
(i) u_i = \(\frac{2\left(x_{i}-0\right)}{1}\), v_i = \(\frac{y_{i}-0}{1}\)
Here, h = 1 and k = 1 are of the same signs.
∴ Corr (U, V) = Corr (X, Y) = 0.8

(ii) u_i = \(\frac{x_{i}-0}{2}\), v_i = \(\frac{y_{\mathrm{i}}-0}{1}\)
Here, h = 2 and k = 1 are of the same signs.
∴ Corr (U, V) = Corr (X, Y) = 0.8

(iii) Corr (X, 3Y) = Corr (X, Y) = 0.8

(iv) Corr (X – 5, Y – 3) = Corr(X, Y) = 0.8

(v) Corr (X + 7, Y + 9) = Corr(X, Y) = 0.8

(vi) Corr(\(\frac{X-5}{7}, \frac{Y-3}{8}\)) = Corr(X, Y) = 0.8

Question 9.
In the calculation of the correlation coefficient between the height and weight of a group of students of a college, one investigator took the measurements in inches and pounds while the other investigator took the measurements in cm. and kg. Will they get the same value of the correlation coefficient or different values? Justify your answer.
Solution:
The coefficient of correlation is a ratio of covariance and standard deviations.
Since covariance and standard deviations are independent of units of measurement.
∴ coefficient of correlation is also independent of units of measurement.
∴ values of coefficient of correlation obtained by first and second investigators are the same.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Correlation Ex 5.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Correlation Ex 5.1

Question 1.
Draw a scatter diagram for the data given below and interpret it.

Solution:

Since all the points lie in a band rising from left to right.
Therefore, there is a positive correlation between the values of X and Y respectively.

Question 2.
For the following data of marks of 7 students in Physics (x) and Mathematics (y), draw scatter diagram and state the type of correlation.

Solution:
We take marks in Physics on X-axis and marks in Mathematics on Y-axis and plot the points as below.

We get a band of points rising from left to right. This indicates the positive correlation between marks in Physics and marks in Mathematics.

Question 3.
Draw a scatter diagram for the data given below. Is there any correlation between Aptitude score and Grade points?

Solution:

The points are completely scattered i.e., no trend is observed.
∴ there is no correlation between Aptitude score (X) and Grade point (Y).

Question 4.
Find correlation coefficient between x andy series for the following data:
n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, σ_x = 3.01, σ_y = 3.03, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 122
Solution:
Here, n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, σ_x = 3.01, σ_y = 3.03, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 122

Question 5.
The correlation coefficient between two variables x and y are 0.48. The covariance is 36 and the variance of x is 16. Find the standard deviation of y.
Solution:
Given, r = 0.48, Cov(X, Y) = 36
Since \(\sigma_{X}^{2}\) = 16
∴ σ_x = 4

∴ the standard deviation of y is 18.75.

Question 6.
In the following data, one of the values of y is missing. Arithmetic means of x and y series are 6 and 8 respectively. (√2 = 1.4142)

(i) Estimate missing observation.
(ii) Calculate correlation coefficient.
Solution:
(i) Let X = x_i, Y = y_i and missing observation be ‘a’.
Given, \(\bar{x}\) = 6, \(\bar{y}\) = 8, n = 5
∴ 8 = \(\frac{35+a}{5}\)
∴ 40 = 35 + a
∴ a = 5
(ii) We construct the following table:

Question 7.
Find correlation coefficient from the following data. [Given: √3 = 1.732]

Solution:

Question 8.
The correlation coefficient between x and y is 0.3 and their covariance is 12. The variance of x is 9, find the standard deviation of y.
Solution:
Given, r = 0.3, Cov(X, Y) = 12,
\(\sigma_{X}^{2}\) = 9
∴ \(\sigma_{\mathrm{X}}\) = 3

∴ the standard deviation of y is 13.33.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Miscellaneous Exercise 4

Question 1.
Following data gives the coded price (X) and demand (Y) of a commodity.

Classify the data by taking classes 0 – 4, 5 – 9, etc. for X and 5 – 8, 9 – 12, etc. for Y.
Also find
(i) marginal frequency distribution of X and Y.
(ii) conditional frequency distribution of Y when X is less than 10.
Solution:
Given, X = coded price
Y = demand
Bivariate frequency table can be prepared by taking class intervals 0 – 4, 5 – 9,… etc for X and 5 – 8, 9 – 12,… etc for Y.
Bivariate frequency distribution is as follows.

(i) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of Y when X < 10:

Question 2.
Following data gives the age in years and marks obtained by 30 students in an intelligence test.


Prepare a bivariate frequency distribution by taking class intervals 16 – 18, 18 – 20,…,etc. for age and 10 – 20, 20 – 30,…, etc. for marks.
Find
(i) marginal frequency distributions.
(ii) conditional frequency distribution of marks obtained when age of students is between 20 – 22.
Solution:
Let X = Age in years
Y = Marks
Bivariate frequency table can be prepared by taking class intervals 16 – 18, 18 – 20,…, etc for X and 10 – 20, 20 – 30,…, etc for Y.
Bivariate frequency distribution is as follows:

(i) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of Y when X is between 20 – 22:

Question 3.
Following data gives Sales (in Lakh ?) and Advertisement Expenditure (in Thousand ₹) of 20 firms.
(115, 61) (120, 60) (128, 61) (121, 63) (137, 62) (139, 62) (143, 63) (117, 65) (126, 64) (141, 65) (140, 65) (153, 64) (129, 67) (130, 66) (150, 67) (148, 66) (130, 69) (138, 68) (155, 69) (172, 68)
(i) Construct a bivariate frequency distribution table for the above data by taking classes 115 – 125, 125 – 135, ….etc. for sales and 60 – 62, 62 – 64, …etc. for advertisement expenditure.
(ii) Find marginal frequency distributions
(iii) Conditional frequency distribution of Sales when the advertisement expenditure is between 64 – 66 (Thousand ₹)
(iv) Conditional frequency distribution of advertisement expenditure when the sales are between 125 – 135 (lakh ₹)
Solution:
(i) Let X = Sales (in lakh ₹)
Y = Advertisement Expenditure (in Thousand ₹)
Bivariate frequency table can be prepared by taking class intervals 115 – 125, 125 – 135, …. etc for X and 60 – 62, 62 – 64, ….etc for Y.
Bivariate frequency distribution is as follows:

(ii) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of X when Y is between 64 – 66:

(iii) Conditional frequency distribution of Y when X is between 125 – 135:

Question 4.
Prepare a bivariate frequency distribution for the following data, taking class intervals for X as 35 – 45, 45 – 55, …. etc and for Y as 115 – 130, 130 – 145, … etc where, X denotes the age in years and Y denotes blood pressure for a group of 24 persons.
(55, 151) (36, 140) (72, 160) (38, 124) (65, 148) (46, 130) (58, 152) (50, 149) (38, 115) (42, 145) (41, 163) (47, 161) (69, 159) (60, 161) (58, 131) (57, 136) (43, 141) (52, 164) (59, 161) (44, 128) (35, 118) (62, 142) (67, 157) (70, 162)
Also find
(i) Marginal frequency distribution of X.
(ii) Conditional frequency distribution of Y when X < 45.
Solution:
Given X = Age in years
Y = Blood pressure
Bivariate frequency table can be prepared by taking class intervals 35 – 45, 45 – 55, …, etc for X and 115 – 130, 130 – 145, ….., etc for Y.
Bivariate frequency distribution is as follows:

(i) Marginal frequency distribution of X:

(ii) Conditional frequency distribution of Y when X < 45:

Question 5.
Thirty pairs of values of two variables X and Y are given below. Form a bivariate frequency table. Also find marginal frequency distributions of X and Y.

Solution:
Bivariate frequency table can be prepared by taking class intervals 80 – 90, 90 – 100, etc for X and 500 – 600, 600 – 700, …., etc for Y.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y

Question 6.
The following table shows how the samples of Mathematics and Economics scores of 25 students are distributed:

Find the value of ϰ² statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{35 \times 25}{50}\) = 17.5
E₁₂ = \(\frac{35 \times 25}{50}\) = 17.5
E₂₁ = \(\frac{15 \times 25}{50}\) = 7.5
E₂₂ = \(\frac{15 \times 25}{50}\) = 7.5
Table of expected frequencies.

Question 7.
Compute ϰ² statistic from the following data:

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{50 \times 60}{100}\) = 30
E₁₂ = \(\frac{50 \times 40}{100}\) = 20
E₂₁ = \(\frac{50 \times 60}{100}\) = 30
E₂₂ = \(\frac{50 \times 40}{100}\) = 20
Table of expected frequencies.

Question 8.
The attitude of 250 employees towards a proposed policy of the company is as observed in the following table. Calculate ϰ² statistic.

Solution:
Table of observed frequencies

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{150 \times 95}{250}\) = 57
E₁₂ = \(\frac{150 \times 95}{250}\) = 57
E₁₃ = \(\frac{150 \times 60}{250}\) = 36
E₂₁ = \(\frac{100 \times 95}{250}\) = 38
E₂₂ = \(\frac{100 \times 95}{250}\) = 38
E₂₃ = \(\frac{100 \times 60}{250}\) = 24
Table of observed frequencies.

Question 9.
In a certain sample of 1000 families, 450 families are consumers of tea. Out of 600 Hindu families, 286 families consume tea. Calculate ϰ² statistic.
Solution:
The given data can be arranged in the following table.

Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{600 \times 450}{1000}\) = 270
E₁₂ = \(\frac{600 \times 550}{1000}\) = 330
E₂₁ = \(\frac{400 \times 450}{1000}\) = 180
E₂₂ = \(\frac{400 \times 550}{1000}\) = 220
Table of expected frequencies.

Question 10.
A sample of boys and girls were asked to choose their favourite sport, with the following results. Find the value of ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{200 \times 126}{300}\) = 84
E₁₂ = \(\frac{200 \times 90}{300}\) = 60
E₁₃ = \(\frac{200 \times 69}{300}\) = 46
E₁₄ = \(\frac{200 \times 15}{300}\) = 10
E₂₁ = \(\frac{100 \times 126}{300}\) = 42
E₂₂ = \(\frac{100 \times 90}{300}\) = 30
E₂₃ = \(\frac{100 \times 69}{300}\) = 23
E₂₄ = \(\frac{100 \times 15}{300}\) = 5
Table of expected frequencies.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.2

Question 1.
The following table shows the classification of applications for secretarial and for sales positions according to gender. Calculate the value of ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{225 \times 100}{300}\) = 75
E₁₂ = \(\frac{225 \times 200}{300}\) = 150
E₂₁ = \(\frac{75 \times 100}{300}\) = 25
E₂₂ = \(\frac{75 \times 200}{300}\) = 50
Table of expected frequencies.

Question 2.
200 teenagers were asked which takeaway food do they prefer – French fries, burgers, or pizza. The results were-

Compute ϰ² statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{50 \times 24}{200}\) = 6
E₁₂ = \(\frac{50 \times 60}{200}\) = 15
E₁₃ = \(\frac{50 \times 116}{200}\) = 29
E₂₁ = \(\frac{150 \times 24}{200}\) = 18
E₂₂ = \(\frac{150 \times 60}{200}\) = 45
E₂₃ = \(\frac{150 \times 116}{200}\) = 87
Table of expected frequencies.

Question 3.
A sample of men and women who had passed their driving test either in 1st attempt or in 2nd attempt
were surveyed. Compute ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{60 \times 40}{80}\) = 30
E₁₂ = \(\frac{60 \times 40}{80}\) = 30
E₂₁ = \(\frac{20 \times 40}{80}\) = 10
E₂₂ = \(\frac{20 \times 40}{80}\) = 10
Table of expected frequencies.

Question 4.
800 people were asked whether they wear glasses for reading with the following results.

Compute the ϰ² square statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{400 \times 600}{800}\) = 300
E₁₂ = \(\frac{400 \times 200}{800}\) = 100
E₂₁ = \(\frac{400 \times 600}{800}\) = 300
E₂₂ = \(\frac{400 \times 200}{800}\) = 100
Table of expected frequencies.

Question 5.
Out of a sample of 120 persons in a village, 80 were administered a new drug for preventing influenza, and out of the 18 were attacked by influenza. Out of those who are not administered the new drug, 10 persons were not attacked by influenza:
(i) Prepare a two-way table showing frequencies.
(ii) Compute the ϰ² square statistic.
Solution:
(i) The given data can be arranged in the following table.

The observed frequency table can be prepared as follows:

(ii) Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{48 \times 80}{120}\) = 32
E₁₂ = \(\frac{48 \times 40}{120}\) = 16
E₂₁ = \(\frac{72 \times 80}{120}\) = 48
E₂₂ = \(\frac{72 \times 40}{120}\) = 24
Table of expected frequencies.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.1

Question 1.
The following table gives income (X) and expenditure (Y) of 25 families:

Find
(i) Marginal frequency distributions of income and expenditure.
(ii) Conditional frequency distribution of X when Y is between 300 – 400.
(iii) Conditional frequency distribution of Y when X is between 200 – 300.
(iv) How many families have their income ₹ 300 and more and expenses ₹ 400 and less?
Solution:
The bivariate frequency distribution table for Income (X) and Expenditure (Y) is as follows:

(i) Marginal frequency distribution of income (X):

Marginal frequency distribution of expenditure ( Y):

(ii) Conditional frequency distribution of X when Y is between 300 – 400:

(iii) Conditional frequency distribution of Y when X is between 200 – 300:

(iv) The cells 300 – 400 and 400 – 500 are having income ₹ 300 and more and the cells 200 – 300 and 300 – 400 are having expenditure ₹ 400 and less.
Now, the following table indicates the number of families satisfying the above condition.

∴ There are 17 families with income ₹ 300 and more and expenditure ₹ 400 and less.

Question 2.
Two dice are thrown simultaneously 25 times. The following pairs of observations are obtained.
(2, 3) (2, 5) (5, 5) (4, 5) (6, 4) (3, 2) (5, 2) (4, 1) (2, 5) (6, 1) (3, 1) (3, 3) (4, 3) (4, 5) (2, 5) (3, 4) (2, 5) (3, 4) (2, 5) (4, 3) (5, 2) (4, 5) (4, 3) (2, 3) (4, 1)
Prepare a bivariate frequency distribution table for the above data. Also, obtain the marginal distributions.
Solution:
Let X = Observation on 1st die
Y = Observation on 2nd die
Now, the minimum value of X is 1 and the maximum value is 6.
Also, the minimum value of Y is 1 and the maximum value is 6.
A bivariate frequency distribution can be prepared by taking X as row and Y as a column.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Question 3.
Following data gives the age of husbands (X) and age of wives (Y) in years. Construct a bivariate frequency distribution table and find the marginal distributions.

Find conditional frequency distribution of age of husbands when the age of wife is 23 years.
Solution:
Given, X = Age of Husbands (in years)
Y = Age of Wives (in years)
Now, the minimum value of X is 25 and the maximum value is 29.
Also, the minimum value of Y is 19 and the maximum value is 23.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of X when Y is 23:

Question 4.
Construct a bivariate frequency distribution table of the marks obtained by students in Statistics (X) and English (Y).

Construct a bivariate frequency distribution table for the above data by taking class intervals 20 – 30, 30 – 40, …. etc. for both X and Y. Also find the marginal distributions and conditional frequency distribution of Y when X lies between 30 – 40.
Solution:
Given, X = Marks in Statistics
Y = Marks in English
A bivariate frequency table can be prepared by taking class intervals 20 – 30, 30 – 40,…, etc for both X and Y.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of Y when X lies between 30 – 40:

Question 5.
Following data gives height in cms (X) and weight in kgs (Y) of 20 boys. Prepare a bivariate frequency table taking class intervals 150 – 154, 155 – 159,…etc. for X and 35 – 39, 40 – 44,…, etc for Y. Also, find
(i) Marginal frequency distributions.
(ii) Conditional frequency distribution of Y when 155 ≤ X ≤ 159.
(152,40) (160,54) (163,52) (150,35) (154,36) (160,49) (166,54) (157,38)
(159,43) (153,48) (152,41) (158,51) (155,44) (156,47) (156,43) (166,53)
(160,50) (151,39) (153,50) (158,46)
Solution:
Given X = Height in cms.
Y = Weight in kgs.
Bivariate frequency table can be prepared by taking class intervals 150 – 154, 155 – 159, …, etc for X and 35 – 39, 40 – 44,…etc for Y.
The bivariate frequency distribution table is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of Y when 155 ≤ X ≤ 159:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Skewness Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Skewness Miscellaneous Exercise 3

Question 1.
For u distribution, mean = 100, mode = 80 and S.D. = 20. Find Pearsonian coefficient of skewness Sk_p.
Solution:
Given, Mean = 100, Mode = 80, S.D. = 20

Question 2.
For a distribution, mean = 60, median = 75 and variance = 900. Find Pearsonian coefficient of skewness Sk_p.
Solution:
Given. Mean = 60, Median = 75, Variance = 900
∴ S.D. = √Variance = √900 = 30

Question 3.
For a distribution, Q₁ = 25, Q₂ = 35 and Q₃ = 50. Find Bowley’s coefficient of skewness Sk_b.
Solution:
Given Q₁ = 25, Q₂ = 35, Q₃ = 50

Question 4.
For a distribution Q₃ – Q₂ = 40, Q₂ – Q₁ = 60. Find Bowlev’s coefficient of skewness Sk_b.
Solution:
Given, Q₃ – Q₂ = 40, Q₂ – Q₁ = 60

Question 5.
For a distribution, Bowley’s coefficient of skewness is 0.6. The sum of upper and lower quartiles is 100 and median is 38. Find the upper and lower quartiles.
Solution:
Given, Sk_b = 0.6, Q₃ + Q₁ = 100,
Median = Q₂ = 38
Skb = \(\frac{Q_{3}+Q_{1}-2 Q_{2}}{Q_{3}-Q_{1}}\)
∴ 0.6 = \(\frac{100-2(38)}{Q_{3}-Q_{1}}\)
∴ 0.6(Q₃ – Q₁) = 100 – 76 = 24
∴ Q₃ – Q₁ = 40 ….(i)
Q₃ + Q₁ = 100 …..(ii) (given)
Adding (i) and (ii), we get
2Q₃ = 140
∴ Q₃ = 70
Substituting the value of Q₃ in (ii), we get
70 + Q₁ = 100
∴ Q₁ = 100 – 70 = 30
∴ upper quartile = 70 and lower quartile = 30

Question 6.
For a frequency distribution, the mean is 200, the coefficient of variation is 8% and Karl Pearsonian’s coefficient of skewness is 0.3. Find the mode and median of the distribution.
Solution:
Mean = \(\bar{x}\) = 200
Coefficient of variation, C.V. = 8%, Skp = 0.3
C.V. = \(\frac{\sigma}{\bar{x}} \times 100\), where σ = standard deviation
∴ 8 = \(\frac{\sigma}{200} \times 100\)
∴ σ = \(\frac{8 \times 200}{100}\) = 16
Now, Sk_p = \(\frac{\text { Mean – Mode }}{\text { S.D. }}\)
∴ 0.3 = \(\frac{200-\text { Mode }}{16}\)
∴ 0.3 × 16 = 200 – Mode
∴ Mode = 200 – 4.8 = 195.2
Since, Mean – Mode = 3(Mean – Median)
∴ 200 – 195.2 = 3(200 – Median)
∴ 4.8 = 600 – 3Median
∴ 3Median = 600 – 4.8 = 595.2
∴ Median = 198.4

Question 7.
Calculate Karl Pearsonian’s coefficient of skewness Skp from the follow ing data:

Solution:
The given table is the cumulative frequency table of more than type.
From this table, we have to prepare the frequency distribution table and then calculate the value of Sk_p.
Construct the following table:

From the table, N = 120, Σf_ix_i = 5490 and \(\sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}^{2}\) = 284600
Mean = \(\bar{x}=\frac{\sum f_{i} x_{i}}{N}=\frac{5490}{120}\) = 45.75
Maximum frequency 42 is of the class 50 – 60
∴ Mode lies in the class 50 – 60
∴ L = 50, f₁ = 42, f₀ = 25, f₂ = 13, h = 10

Alternate Method:
Let u = \(\frac{x-45}{10}\)

\(\overline{\mathrm{u}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\mathrm{N}}=\frac{9}{120}\) = 0.075
∴ \(\bar{x}\) = 45 + 10(\(\bar{u}\))
= 45 + 10(0.075)
= 45 + 0.75
= 45.75
Var(u) = \(\sigma_{u}^{2}=\frac{\sum f_{i} u_{i}{ }^{2}}{N}-(\bar{u})^{2}\)
= \(\frac{335}{120}\) – (0.075)²
= 2.7917 – 0.0056
= 2.7861
Var(X) = h² × Var(u)
= 100 × 2.7861
= 278.61
S.D. = √278.61 = 16.6916
Maximum frequency 42 is of the class 50 – 60.
∴ Mode lies in the class 50 – 60.
∴ L = 50, f₁ = 42, f₀ = 25, f₂ = 13, h = 10

Question 8.
Calculate Bowley’s coefficient of skewness Skb from the following data.

Solution:
To calculate Bowley’s coefficient of skewness Skb, we construct the following table:

Here, N = 120
Q₁ class = class containing the \(\left(\frac{N}{4}\right)^{t h}\) observation
∴ \(\frac{N}{4}=\frac{120}{4}\) = 30
Cumulative frequency which is just greater than (or equal to) 30 is 35.
∴ Q₁ lies in the class 30-40.
∴ L = 30, h = 10, f = 13, c.f. = 22

Q₂ class = class containing the \(\left(\frac{N}{2}\right)^{t h}\) observation
∴ \(\frac{\mathrm{N}}{2}=\frac{120}{2}\) = 60
Cumulative frequency which is just greater than (or equal to) 60 is 60.
∴ Q₂ lies in the class 40-50.
∴ L = 40, h = 10, f = 25, c.f. = 35

Q₃ class = class containing the \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 120}{4}\) = 90
Cumulative frequency which is just greater than (or equal to) 90 is 102.
∴ Q₃ lies in the class 50 – 60
∴ L = 50, h = 10, f = 42, c.f. = 60

Question 9.
Find Sk_p for the following set of observations:
18, 27, 10, 25, 31, 13, 28
Solution:
The given data can be arranged in ascending order as follows:
10, 13, 18, 25, 27, 28, 31
Here, n = 7
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= value of \(\left(\frac{7+1}{2}\right)^{\text {th }}\) observation
= value of 4th observation
= 25
For finding standard deviation, we construct the following table:

Question 10.
Find Skb for the following set of observations:
18, 27, 10, 25, 31, 13, 28
Solution:
The given data can be arranged in ascending order as follows:
10, 13, 18, 25, 27, 28, 31
Here, n = 7
∴ Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation
= value of 2nd observation
∴ Q₁ = 13
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 2)th observation
= value of 4th observation
∴ Q₂ = 25
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 2)th observation
= value of 6th observation
∴ Q₃ = 28
Coefficient of skewness,

∴ Sk_b = -0.6