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31 Dec 2015, 03:33
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (03:54) correct
62%
(02:57)
wrong
based on 42
sessions
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There are 5 women and 4 men including 2 couples, Mr. & Mrs. X and Mr. & Mrs. Y, in a club. In how many ways can a committee of 5 members be formed if no married couple can be there in the committee?
a)30 ways
b)51 ways
c)70 ways
d)126 ways
e)131 ways
My answer : 5 member can be formed in 60+40 = 100 ways [(5c3*4c2)+(5c2*4c3)]
2 couples can arrange them in 4! ways = 24 and they can re align them in 2! ways so total 48 ways
therefore 100-48 =52 ways but the answer choices are different
Please provide explanations if you have?
a)30 ways
b)51 ways
c)70 ways
d)126 ways
e)131 ways
My answer : 5 member can be formed in 60+40 = 100 ways [(5c3*4c2)+(5c2*4c3)]
2 couples can arrange them in 4! ways = 24 and they can re align them in 2! ways so total 48 ways
therefore 100-48 =52 ways but the answer choices are different
Please provide explanations if you have?
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31 Dec 2015, 04:46
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jyotibrata
Hi,
where you are going wrong is :-
1) why are you excluding the ways when all 5 are women, or when 4 are women and 1 is men...
2) this is forming committee and arrangement does not matter, so it is combination and not permutation so 4! etc are not required..
the answer which I have got does not seem to be in answer, pl relook at the answer..
31 Dec 2015, 07:16
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Agreed chetan2u. Even the answer which I obtained is not in the answer choice. Let me know if my approach is correct.
Total = 9 members
Out of 9 members, 4 are married and 5 are unmarried.
Number of ways of selecting 5 members so that no married couples are there together = 5C3 * 2C1 * 2C1 + 5C5 * 2C0 * 2C0 = 40 + 1 = 41 ways
5C3 * 2C1 * 2C1 --> 3 in the committee are unmarried and 1 each from married 2 couples.
5C5 * 2C0 * 2C0 --> all 5 in the committee are unmarried.
Total = 9 members
Out of 9 members, 4 are married and 5 are unmarried.
Number of ways of selecting 5 members so that no married couples are there together = 5C3 * 2C1 * 2C1 + 5C5 * 2C0 * 2C0 = 40 + 1 = 41 ways
5C3 * 2C1 * 2C1 --> 3 in the committee are unmarried and 1 each from married 2 couples.
5C5 * 2C0 * 2C0 --> all 5 in the committee are unmarried.
