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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

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...

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.

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Re: If there are four distinct pairs of brothers and sisters, th [#permalink]
06 Jun 2014, 02:01

Expert's post

4

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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?

A. 8 B. 24 C. 32 D. 56 E. 192

As committee shouldn't have siblings in it, then a pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters should send one "representative" to the committee is \(C^3_4\) (choosing 3 pairs which will be granted the right to send one "representative" to the committee);

But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: \(2*2*2=2^3\);

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