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Permutations
and Combinations
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1.
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10. Is the given statement true or
false?
nCr= nCn-r
a) True
b) False
Answer: a
Explanation: The property of combination states nCr= nCn-r
As nCr=n!(n−r)!×r!
nCn−r=n!r!×(n−r)!= nCr.
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A
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2.
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Number of circular permutations of
different things taken all at a time is n!.
a) True
b) False
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!.
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B
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3.
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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
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.
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C
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4.
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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
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 = nC3
– mC3
Therefore the number of triangles = 20C3 – 5C3
= 1140-10 = 1130.
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A
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5.
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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
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.
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B
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6.
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If nPr =
3024 and nCr = 126 then find n and r.
a) 9, 4
b) 10, 3
c) 12, 4
d) 11, 4
Answer: a
Explanation: nPrnCr=3024126
nPr=n!(n−r)!
nCr=n!(n−r)!×r!
Hence [n!(n−r)!]÷[n!(n−r)!×r!] = 24
24 = r!
Hence r = 4
Now nP4 = 3024
n!(n−4)!=3024
n(n-1)(n-2)(n-3) = 9.8.7.6
n = 9.
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A
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7.
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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
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.
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B
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8.
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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
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.
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C
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9.
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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
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.
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C
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10.
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If 16Pr-1 : 15Pr-1
= 16 : 7 then find r.
a) 10
b) 12
c) 7
d) 8
Answer: a
Explanation: We know that nPr=n!(n−r)!
Hence 16Pr−1:15Pr−1=16:7
[16!16−(r−1)!]÷[15!15−(r−1)]=16÷7
16 ÷ (17 – r) = 16 ÷ 7
17 – r = 7
Hence r = 10.
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A
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