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M02-05

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Intern
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Joined: 28 Aug 2016
Posts: 44
Re: M02-05  [#permalink]

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New post 07 Jun 2018, 19:19
Bunuel wrote:
shohm wrote:
Please help understand where my thinking is flawed ?
A 3 member committee needs to be formed from 4 pair of distinct siblings say (b1,s1), (b2, s2), (b3, s3), (b4,s4).

1st member can be any of the 8 persons. Now, for next member selection, discard the pair from which the 1st member was selected. So now we have 3 pairs left to choose from.
2nd member can be any of the 6 persons in the left 3 pairs. Now, for next member selection, discard the pair from which the 1st and 2nd member were selected. So now we have 2 pairs left to choose from.
3rd member can be any of the 4 persons in the remaining 2 pairs.

So total number of ways = 8*6*4 = 192. Where am I counting the extras ?


Check here: http://gmatclub.com/forum/if-there-are- ... ml#p775925

Hope it helps.



Can anyone explain why the below is wrong?

To calculate the numbers of cases that have siblings, we select 2 pairs out of 4 pairs (1 pair included in the committee and only 1 person of the 2nd pair in the committee): 4C2 *2 = 12

Why is this wrong?

If we are calculating the cases with siblings that that means those 3 people belong to 2 pairs, and the 3rd person can either be sister or a brother
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Re: M02-05  [#permalink]

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New post 15 Jun 2018, 00:34
3C8 - total number of ways to select committee of 3 people = 56
4! = 24 number of ways to select sibling pairs.
56-24 = 32
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Re: M02-05  [#permalink]

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New post 18 Jul 2018, 06:41
why is my approach wrong ?

4C3 + 4C3 + (4C2 * 3C1) + (4C2 * 3C1)

selecting all brothers + all sisters + 2 brothers and 1 sister of 3 + 2 sisters and 1 brother of 3.

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Re: M02-05 &nbs [#permalink] 18 Jul 2018, 06:41

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