Probability and Statistics Questions and Answers – Permutations and Combinations

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

1. If 16Pr-1 : 15Pr-1 = 16 : 7 then find r.
a) 10
b) 12
c) 7
d) 8
View Answer

Answer: a
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.
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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

Answer: c
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

Answer: c
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 3P2 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

Answer: b
Explanation: The given digits are 1, 3, 5, 7, 9
Sum of r digit number= n-1Pr-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
4P3 (1+3+5+7+9)(1111)=666600.

5. If nPr = 3024 and nCr = 126 then find n and r.
a) 9, 4
b) 10, 3
c) 12, 4
d) 11, 4
View Answer

Answer: a
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 nP4 = 3024
\(\frac{n!}{(n-4)!} = 3024 \)
n(n-1)(n-2)(n-3) = 9.8.7.6
n = 9.
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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

Answer: b
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 = 9C2 × 9C2 = 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 = 12+22…..82 = 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

Answer: a
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 = nC3mC3
Therefore the number of triangles = 20C35C3 = 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) 10C3
b) 9C5
c) 9C6
d) 9C4
View Answer

Answer: c
Explanation: One particular person is not included we have to select 6 persons out of 9 which can be done in 9C6 ways.

9. Number of circular permutations of different things taken all at a time is n!.
a) True
b) False
View Answer

Answer: b
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?
nCr= nCn-r
a) True
b) False
View Answer

Answer: a
Explanation: The property of combination states nCr= nCn-r
As \(^nC_r = \frac{n!}{(n-r)!×r!} \)
\(^nC_{n-r} = \frac{n!}{r!×(n-r)!} = ^nC_r. \)
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn