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A group consists of 17 men and 22 women. How many pairs can

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New post 07 Nov 2004, 15:29
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A group consists of 17 men and 22 women. How many pairs can be made to contain at least one woman?
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New post 07 Nov 2004, 15:41
# of ways to choose 1 man and 1 woman = 17C1*22C1=374
# of ways to choose 0 man and 2 women = 17C0*22C2= 231

answer = 374+231 = 605
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New post 07 Nov 2004, 18:15
Do we consider each man and woman a unique identity in such a question?

If this is not the case then the solution will be simply 19
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New post 07 Nov 2004, 18:46
obada wrote:
Do we consider each man and woman a unique identity in such a question?

If this is not the case then the solution will be simply 19

How can a person not be unique :)
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New post 07 Nov 2004, 19:10
17C1*22C1 + 17C0*22C2 = whatever this may be.
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New post 08 Nov 2004, 03:44
To find the no. of pairs consisting of atleast 1 women, find total possible no. of pairs i.e. (39C2) and subtract it from the total no of pairs that are both males. i.e. (17C2)

Hence the answer is 39C2 - 17C2 = 741 - 136 = 605

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New post 08 Nov 2004, 12:53
We can have a pair of two women or 1 man and 1 woman.
2 women can be selected in 22C2 ways=231
and 1 man and woman in 17C1*22C1 ways=374

total combination= 231+374=605.
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New post 08 Nov 2004, 19:48
OA is 605 :good
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  [#permalink] 08 Nov 2004, 19:48
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