Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: If there are four distinct pairs of brothers and sisters, th [#permalink]
06 Jun 2014, 02:01
This post was BOOKMARKED
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;