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There are five men and five women in the room. In how many ways can they be paired-up?

How should I approach this question?

The answer I have in my notes is 5!=120 ways which I don't quite understand.

This is my thought process: 1. Choosing a pair at a time. 2. 1st pair, the number of ways they can be paired-up = 5 * 5 = 25 ways 3. 2nd pair, 4 * 4 = 16 4. 3rd pair, 3 * 3 = 9 5. 4th pair, 2 * 2 = 4 6. 5th pair, 1 * 1 = 1 Total Number of ways they can be paired = 55 ways.

Anlin yang
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  • To clarify your intuition, it might help to start with a smaller collection. Easy if we replace $5$ by $1,2$. Working the case of $3$ should help. – lulu Jun 11 '17 at 12:55

2 Answers2

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Hint. Instead of choosing "a pair at a time" think about lining the men up (in any fixed order) and then choosing a woman for each, "one at a time". That means $5$ ways for the first man, $4$ for the second, and so on. Try it out (perhaps just for three couples) and you'll see why these values should be multiplied, not added.

Ethan Bolker
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First, in your approach you should not sum up the possibilities but multiply them (for a pairing, you have 25 choices for the first pair, 16 for the second pair and so on). In any case this gives you too many pairings as you have counted pairings multiple times, you need to "cancel out" the permutations. As you have 5 pairs, you have 5! ways to permute them. So with your approach you get

$$\frac{5*5*4*4*3*3*2*2*1*1}{5*4*3*2*1}=5!$$ pairings.

you could also fix one permutation of the men. The first man then can choose from 5 women, the second one from 4,...., so using this you also get 5! pairings

sincerely

slinshady

Pratyush Yadav
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slinshady
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