This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Permutations and Combinations”.

1. If ^{16}P_{r-1} : ^{15}P_{r-1} = 16 : 7 then find r.

a) 10

b) 12

c) 7

d) 8

View Answer

Explanation: We know that \(^nP_r = \frac{n!}{(n-r)!} \)

Hence \(^{16}P_{r-1} : \, ^{15}P_{r-1} = 16 : 7 \)

\([\frac{16!}{16-(r-1)!}] ÷ [\frac{15!}{15-(r-1)}] = 16 ÷ 7 \)

16 ÷ (17 – r) = 16 ÷ 7

17 – r = 7

Hence r = 10.

2. In a colony, there are 55 members. Every member posts a greeting card to all the members. How many greeting cards were posted by them?

a) 990

b) 890

c) 2970

d) 1980

View Answer

Explanation: First player can post greeting cards to the remaining 54 players in 54 ways. Second player can post greeting card to the 54 players. Similarly, it happens with the rest of the players. The total numbers of greeting cards posted are

54 + 54 + 54 …

54 (55times) = 54 x 55 = 2970.

3. Find the number of ways of arranging the letters of the words DANGER, so that no vowel occupies odd place.

a) 36

b) 48

c) 144

d) 96

View Answer

Explanation: The given word is DANGER. Number of letters is 6. Number of vowels is 2 (i.e., A, E). Number of consonants is 4 (i.e., D,N,G,R). As the vowels cannot occupy odd places, they can be arranged in even places. Two vowels can be arranged in 3 even places in

^{3}P

_{2}ways i.e., 6. Rest of the consonants can arrange in the remaining 4 places in 4! ways. The total number of arrangements is 6 x 4! = 144.

4. Find the sum of all four digit numbers that can be formed by the digits 1, 3, 5, 7, 9 without repetition.

a) 666700

b) 666600

c) 678860

d) 665500

View Answer

Explanation: The given digits are 1, 3, 5, 7, 9

Sum of r digit number=

^{n-1}P

_{r-1}

(Sum of all n digits)×(1111… r times)

N is the number of non zero digits.

Here n=5, r=4

The sum of 4 digit numbers

^{4}P

_{3}(1+3+5+7+9)(1111)=666600.

5. If ^{n}P_{r} = 3024 and ^{n}C_{r} = 126 then find n and r.

a) 9, 4

b) 10, 3

c) 12, 4

d) 11, 4

View Answer

Explanation: \(\frac{^nP_r}{^nC_r} = \frac{3024}{126} \)

\(^nP_r = \frac{n!}{(n-r)!} \)

\(^nC_r = \frac{n!}{(n-r)!×r!} \)

Hence \( [\frac{n!}{(n-r)!}]÷[\frac{n!}{(n-r)!×r!}] \) = 24

24 = r!

Hence r = 4

Now

^{n}P

_{4}= 3024

\(\frac{n!}{(n-4)!} = 3024 \)

n(n-1)(n-2)(n-3) = 9.8.7.6

n = 9.

6. Find the number of rectangles and squares in an 8 by 8 chess board respectively.

a) 296, 204

b) 1092, 204

c) 204, 1092

d) 204, 1296

View Answer

Explanation: Chess board consists of 9 horizontal 9 vertical lines. A rectangle can be formed by any two horizontal and two vertical lines. Number of rectangles =

^{9}C

_{2}×

^{9}C

_{2}= 1296. For squares there is one 8 by 8 square four 7 by 7 squares, nine 6 by 6 squares and like this

Number of squares on chess board = 1

^{2}+2

^{2}…..8

^{2}= 204

Only rectangles = 1296-204 = 1092.

7. There are 20 points in a plane, how many triangles can be formed by these points if 5 are colinear?

a) 1130

b) 550

c) 1129

d) 1140

View Answer

Explanation: Number of points in plane n = 20.

Number of colinear points m = 5.

Number of triangles from by joining n points of which m are colinear =

^{n}C

_{3}–

^{m}C

_{3}

Therefore the number of triangles =

^{20}C

_{3}–

^{5}C

_{3}= 1140-10 = 1130.

8. In how many ways can we select 6 people out of 10, of which a particular person is not included?

a) ^{10}C_{3}

b) ^{9}C_{5}

c) ^{9}C_{6}

d) ^{9}C_{4}

View Answer

Explanation: One particular person is not included we have to select 6 persons out of 9 which can be done in

^{9}C

_{6}ways.

9. Number of circular permutations of different things taken all at a time is n!.

a) True

b) False

View Answer

Explanation: The number of circular permutations of different things taken all at a time is n-1! and the number of linear permutations of different things taken all at a time is n!.

10. Is the given statement true or false?

^{n}C_{r}= ^{n}C_{n-r}

a) True

b) False

View Answer

Explanation: The property of combination states

^{n}C

_{r}=

^{n}C

_{n-r}

As \(^nC_r = \frac{n!}{(n-r)!×r!} \)

\(^nC_{n-r} = \frac{n!}{r!×(n-r)!} = ^nC_r. \)

**Sanfoundry Global Education & Learning Series – Probability and Statistics.**

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