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Permutation and Combination problem

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by priyasaibaba » Wed Jun 02, 2010 4:56 pm
Can anyone help to solve this problem?


A 3-people committee is to be selected from four couples. If the committee cannot contain a couple at the same time, how many such committees are possible?
A) 21
(B) 35
(C) 28
(D) 24
(E) 32

OA is E
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by indiantiger » Wed Jun 02, 2010 5:22 pm
Total possible cases : 8C3 = 56

Total cases with a couple =

Position one can be filled in 4 ways (we can pick one person from any four of the couple)
position 2 can be filled in one way ( to make in a couple)
position 3 can be filled with 6 ways ( 2 people have already been taken out of 8)

=> 4*6 = 24

total ways w/o a couple in the committee = 56 - 24 = 32 (E)

Please do correct me if I am wrong.
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by tpr-becky » Thu Jun 03, 2010 12:02 am
This is a combination problem so you have to figure out how many orders are possible by determining how many people are available for each slot and then multiplying them but becuase order does not matter you have to divide out all of the orders which will contain the same people - for example ABC and BCA are different orders but they are the same group of people when order doesn't matter. The easiest way to think of this is to divide by the number of slots factorial.

in the first slot you can pick any of the 8 people - so 8 possibilities.

in the second slot there are only 6 people - you already chose one and you can't pick the one from the same couple.

For the last slot there are 4 people available - from the last two couples.

So 8*6*4/(3*2*1) - for a total of 32.
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by saurabhmahajan » Thu Jun 03, 2010 6:08 am
Friends,
Please let me know if my way is correct of solving.

3 out of 8,so 8C3=56
no couple together,so out of 4 couple there are 4 ways to select a couple,so 4!=24
therefore 56-24=32 (E)
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by tpr-becky » Thu Jun 03, 2010 6:22 am
yes, the alternate way is correct as well. There is often more than one way to do a problem.
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by Thouraya » Wed Jun 09, 2010 5:56 am
TPR,

Why did u divide by 6 in the last step?

Thanks
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by tpr-becky » Wed Jun 09, 2010 6:33 am
Because the problem is a combination (no ordering of the elements) and when that happens you have to divide out all the extra ways. for example if you have AB you wouldn't need BA becuase they would be the same group - and you would divide the 2 ways by 2 to get one way. In other words you divide by the number of ways you can arrange the people in the group.

This is the same when you have three things in a group except the ways you can arrange them is determined by 3! - or 3*2*1=6
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