# How many ways to form a committee, subject to certain restrictions?

From among six couples, a committee of $5$ members is to be formed. If the selected committee has no couple, then in how many ways can the committee be formed?

My approach is to count things step by step like first compute the number of ways such that there is no male, then number of ways there is only one male and his partner is not, two males and their partners are not ... to number of ways there are no males on the committee.

Put in mathematics which looks like: $$\binom{6}{0} \times \binom{6}{1}+\binom{6}{1} \times \binom{5}{4}+\binom{6}{2} \times \binom{4}{3}+ \cdots +\binom{6}{5} \times \binom{1}{0} = 192$$

The answer I get by this approach is correct. However, there is an approach mentioned in my module which counts this whole thing in just $\binom{6}{5} \times 2^5 = 192$. But unfortunately I couldn't understand this approach. Could anybody explain ?

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I don't know what the module does. But note that your five people will be taken from different couples - so first choose 5 couples from 6. Then for each of the five couples you can choose either partner, which gives the $2^5$ factor.

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Choose which couples the committee has a member from,
then choose which spouse from each of those couples is in the committee.

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The expression can be explained in words as

Choose five of the six couples and then from each couple pick a person.

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5 slots to be filled with 12 members.

first slot can be filled in 12 ways. second slot can be filled in 10 ways. (as the partner of first cannot be selected) third slot can be filled in 8 ways. (as the partners of previous members cannot be selected) fourth slot can be filled in 6 ways. (as the partners of previous members cannot be selected) fifth slot can be filled in 4 ways. (as the partners of previous members cannot be selected)

total 12*10*8*6*4 = 23040 ways

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Idea is good, but at the end we must divide by $5!$ since order does not matter. –  André Nicolas Aug 25 '11 at 10:45