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# gc test 2

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Senior Manager
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gc test 2 [#permalink]  31 Jul 2009, 22:47
00:00

Difficulty:

5% (low)

Question Stats:

33% (00:00) correct 66% (00:46) wrong based on 3 sessions
If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

Oa is
[Reveal] Spoiler:
c

pl explain
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Re: gc test 2 [#permalink]  01 Aug 2009, 01:46
vaivish1723 wrote:
If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

Oa is
[Reveal] Spoiler:
c

pl explain

Total no. of combinations = 8C3= 56
Now lets take the case when we have one sibling in the committee. Say 1 pair then the no. of combination 6C1 X 4= 24 ( as we have 4 pair of siblings)

No. of ways when we don't have siblings in it = 56-24= 32
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Re: gc test 2 [#permalink]  01 Aug 2009, 04:45
4c3 for selececting 3 couples then for every three v have 2 chioces
4c3*2*2*2
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Re: gc test 2 [#permalink]  26 Jan 2010, 04:47
Why are we multiplying 4 pair of siblings by 6C1?
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Re: gc test 2 [#permalink]  26 Jan 2010, 09:56
bhumika wrote:
Why are we multiplying 4 pair of siblings by 6C1?

We are trying to calculate the no. of ways in which there is one pair and one other.

so there are 6C1 ways of choosing one guy who can be paired with one pair.

Now there are 4 pairs and total no. of ways = 6C1 x 4

I hope that helps..!
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Re: gc test 2 [#permalink]  29 Jan 2010, 21:51
keeping it simple (b1,s1),(b2,s2),(b3,s3),(b4,s4),
1 Pair produces 2 possible wasy between (b,s) with total picks is 3; 2*2*2
Then amount 4 unique pairs it can for another 4 times for unique pairing 8*4=32... brings you back to combinatorics... being a quicker calculation then the above logic...
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Re: gc test 2 [#permalink]  08 May 2011, 06:15
First find the total number of combinations without any constraints, which is 8c3 = 56 (since we're looking to make a committee of 3 people out of 8 and order doesn't matter).

Then, find out all the ways in which you would have a sibling on the committee. Let's look at one sibling pear (Brother, Sister). The number of ways they can both get on the panel is 2c2 * 6 (the six represents the 3rd person on the committee, as there are 6 people left to choose from for the last spot), which gets you 6 combinations. Multiply that 6 by 4 to incorporate the 4 different pairs of siblings.

Using the info we've calculated, the total number of combinations is 56 - 24 = 32.
Re: gc test 2   [#permalink] 08 May 2011, 06:15
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