permutations siblings

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venmic: Master | Next Rank: 500 Posts; Posts: 222; Joined: Mon Oct 13, 2008 4:04 pmFollowed by: 2 members; Thanked: 3 times

by venmic » Thu Sep 01, 2011 6:09 pm

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

What is the logic in this question

56

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Source: Beat The GMAT — Problem Solving | Thu Sep 01, 2011 6:09 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Sep 01, 2011 8:08 pm

venmic 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

What is the logic in this question

56

Good = Total - Bad.

Total:
Number of combinations of 3 that can be formed from 8 people = 8C3 = 56.

Bad:
A bad committee combines one of the 4 sibling pairs with one of the 6 remaining people.
Number of options for the sibling pair = 4.
Number of options for the third committee member = 6.
To combine these options, we multiply:
Bad committees = 4*6 = 24.

Good committees = total-bad = 56-24 = 32.

The correct answer is C.

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saketk: Legendary Member; Posts: 608; Joined: Sun Jun 19, 2011 11:16 amFollowed by: 8 members; Thanked: 37 times

by saketk » Fri Sep 02, 2011 12:40 am

venmic 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

What is the logic in this question

56

Hi --I think the answer you provided is Wrong. Read the post...

This question is similar to the one of the MGMAT question discussed on the Manhattan forum.. That question was about choosing 3 cars each of different color.

Here is the link -- https://www.manhattangmat.com/forums/the ... 12659.html

Now, Lets look at this question... First make 4 pairs

B1S1 B2S2 B3S3 B4S4

Select the first person from the group of 8 people ( 4 pair = 4*2)

This can be done in 8 WAYS, say B1 was selected.

Now, for the next selection S1 cannot participate ..(Reason -- constraint given in the question says that siblings cannot be together)

Therefore, we have only 6 people to choose from.
2nd person can be chosen in 6 WAYS.

Similarly, 3rd person can be chosen in 4 ways..

Total = 8*6*4 -- is this the answer? [WAIT]

We have to see that the ORDER of selection does not matter. So we divide the product by 3! [3! because we chose 3 people]

The correct answer to this question will be -- [spoiler](8*6*4)/3! = 32[/spoiler]

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