Permutation and Combination Help!

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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.

Soln: No. of committees= 10C4-4C3=207

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

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.

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

Hope this helps!

Damn

Stupid me for not reading the problem properly

Thank you gmatpill!

Pls move this to PS Quant forum.

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.

In the first approach I have some doubts
I understood the logic and cases but in the case of 1 married couple included I am doing: 3c1*2c2* 4c2
3c1=no of ways to choose 1 couple out of 3
4c2=no of ways to choose 2 single out of 4
2c2=choosing both persons from the couple selected

Can you explain where I am wrong?

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