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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.
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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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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