Clarification

007math

Six married couples attend the marriage enrichment planning meeting. How many committees of five can be chosen if no committee is to include a husband and wife team?

I think the answer is $384$ , but the answer key claims it is $792$ .

My work is this: $\binom{6}{5}(2 \times 2 \times 2 \times 2 \times 2 \times 2) = 384$ , and I think its right, so why is it 792 (if it is, the key could be wrong)?

**Posted:** Tue Nov 03, 2009 3:49 pm

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$\mathfrak{\sim 007math\sim }$

Thunder365

Yeah, Im pretty sure the answer is not 792. 792=12C5, which doesnt specify the restrictions.

So think of the problem like this:
A B C D E F.......men
A B C D E F.......women

Where two letters of opposite genders cant be in the same set.

If we considered A B C D E F to be just one person, we have 6C5=6 ways to choose a person.
But they arent the same person. Once we have chosen the 5 letters to be in a group, wlog lets say we chose A B C D E, we actually have the following arrangement:
A B C D E.......men
A B C D E.......women

And there are 2x2x2x2x2 ways to choose whether we want a man or woman.

So Im getting the answer of $\binom{6}{5}(2\times 2\times 2\times 2\times 2) = 192$

But then there is a 99% chance Im wrong, so can anyone confirm this?

**Posted:** Tue Nov 03, 2009 5:09 pm

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randomguy64

I confirm Thunder's solution. 007math, I think you misread the answer key; after all, the last 2 digits are correct, and 7 can be easily mistaken as 1.

**Posted:** Tue Nov 03, 2009 5:20 pm

007math

Oh, OK. I get it know. The test this was taken from was multiple choice and all answer choices were within 200 of 792, so I don't think its a typo. I guess the writers forgot to think about the restrictions when coming up with the answer. Anyway, thanks!

**Posted:** Tue Nov 03, 2009 5:46 pm

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randomguy64

That would make sense, because the answer without restrictions is 792. Interesting.

**Posted:** Tue Nov 03, 2009 7:55 pm