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Hi, I'm new to this site so forgive me if I violate any rules.

I need help with a P&C question:

A group of 10 people consists of 3 married couples and 4 single men. A committee of 4 is to be fo from the 10 people. How many different committees can be formed if the committee can consist of at most 1 married couple?

Would appreciate it if any one of you can reply asap! I need urgent help

Re: Permutation and Combination Help! [#permalink]

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17 Mar 2011, 14:56

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Hi yogibearsayshi,

This video might help you understand the breakdown of this "nested GMAT combination" question.

Entwistle --good job breaking down the problem and doing the problem! But watch out! Your calculations were great, but looks like you forgot to get back to the original question and note that they only care about combinations that involve AT MOST 1 couple. I know--GMAT guys do that a lot. I almost missed it too! They make you do this complicated calculation that makes you forget what specific condition they restrict the problem to.

This video might help you understand combinations and permutations in general.

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18 Mar 2011, 09:14

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Ah--in that case, the question relies on a critical assumption: that couples don't necessarily stay together.

The analysis above assumes that couples stay together.

But if couples do not stay together, then a lot more possibilities exist:

You can have 1 married person, and 3 singles. You can have 2 married people, and 2 singles. You can have 3 married people, and 1 single. You can have 4 married people (only 2 of them can be a couple, the other 2 cannot).

So this complicates things further. In this case, using the method maliyeci and yogibearsayshi suggested above, you find the total possible combinations and then subtract the 1 case scenario that is prohibited.

So you treat the married people like singles in terms of calculations, but then subtract out the case when there are 2 married couples.

So yes, that would be 10C4 - 3C2. 10C4 because you are selecting from 10 and choosing 4 people (regardless of married/single status). 3C2 because you want the # of people possibilities for 2 married couples that you want to subtract from the overall.

This would give you: 10C4 = 10! / (4! * 6!) = (10*9*8*7) / (4*3*2*1) = (10*3*7) / (1) = 210

3C2 = 3! / (2! * 1*) = 3

210 - 3 = 207

I'm not so sure you'd see a question worded like this on the actual GMAT. The calculation with the factorials is a little bit more in depth than I would expect. But good to go through though. _________________

Re: Permutation and Combination Help! [#permalink]

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19 Mar 2011, 12:38

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I 100% DON'T agree with GMAT Pill Instructor initial assumption!! No where in the question it says you can assume married couple don't split up.

*Here is the solution in simple terms:* 1. All 4 single = 4C4 = 1 2. 1 married couple included = 3C1* (for the rest 2 there are 3 cases: 2 are single + 1 is married*1 is single + 2 are married) = 3C1* ( 4C2 + (2C1*2)*4C1 + (2C2*2*2) ) = 3 * (6 + 16 + 4) = 3 * 26 = 78 3. 1 married & 3 single = (3C1*2)*4C3 = 24 4. 2 married & 2 single = (3C2*2*2)*4C2 = 12*6=72 5. 3 married & 1 single = (3C3*2*2*2)*4C1 = 8*4=32

Total = 1+ 78 + 24 + 72 + 32 = 207

I might have done some calculation mistake (forgive me for that), but hope you got the point.

*Alternative approach:* (I always thing of negative approach as if you go to harder levels, GMAT I think likes to test this way of your thinking) At-most 1 married couple = Total possible ways - 2 married couple = 10C4 - 3C2 = 210 - 3 = 207

Re: Permutation and Combination Help! [#permalink]

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18 Mar 2011, 01:58

Hey thanks for the help! And I'll take note of that, Entwistle.

However the answer is supposed to be 207. /: I finally figured out the solution, but it doesn't make sense! I used your method to tackle this question as well.

Re: Permutation and Combination Help! [#permalink]

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18 Mar 2011, 08:08

4 people can be chosen from 10 people in 10C4 = 210 ways. Question only wants us to exclude the last case, 2 couples are in commitee. That is 3C2=3 possibilities. 210-3=207

Re: Permutation and Combination Help! [#permalink]

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04 Apr 2011, 10:11

Expert's post

aznboi986 wrote:

Typically, how many P&C questions will you encounter?

Well, the test is adaptive. P&C questions are usually tougher so if you don't answer enough hard questions to get here, then you might not encounter any. However, if you are looking to score in the 700 range, you will likely encounter a few (2 or 3) in varying difficulties. Often times you can think through the variations, but understanding the formula as well as Permutations vs combinations vs variations will help you answer the question quickly and accurately without having a panic during exam time. _________________

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