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This is a question from Sheldon Ross.

A dance class consists of $22$ students, $10$ women and $12$ men. If $5$ men and $5$ women are to be chosen and then paired off, how many results are possible?

So my reasoning is this :

  1. I choose 5 women from a pool of 10 in $^{10}C_2$ ways.
  2. I choose 5 men from a pool of 12 in $^{12}C_2$ ways.

So total number of ways of choosing in 10C2 x 12C2. Now I need to arrange them in 5 pairs. This is where I have a different solution. The solution says that there are 5! ways to arrange them in pairs.

But I cant seem to understand why? My reasoning is that for first pair position I need to choose 1 man from 5 and 1 woman from 5. So for the first position I have 5 x 5 choices (5 for man and 5 for woman). Similarly for the second position I have 4 x 4 choices and so on. Hence the total ways are 5! x 5!

So I calculate the total ways as $^{10}C_2 * ^{12}C_2 * 5! * 5!$. Can anyone point the flaw in my reasoning for arranging the chosen men and women in pairs.

Thanks in advance

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  • $\begingroup$ Please make your titles informative. We get dozens of "permutations combinations" questions everyday. $\endgroup$
    – Em.
    Commented Jul 10, 2016 at 8:28
  • $\begingroup$ Well, to pair up, the first B/G can choose 5G/B, then 4 and so on, thus it is 5!, not $5!^2$ $\endgroup$
    – Ariana
    Commented Jul 10, 2016 at 8:53
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    $\begingroup$ Did you mean $\binom{10}{5}\binom{12}{5}5!$? $\endgroup$ Commented Jul 10, 2016 at 9:08
  • $\begingroup$ @N.F.Taussig, Yes that's what I meant. That's the answer in book. My reasoning has one more 5!. $\endgroup$
    – amrx
    Commented Jul 10, 2016 at 9:17
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    $\begingroup$ As Mike indicated in his answer, once you have selected five men and five women, you can line up the men in some order, say alphabetically. The first man in the list can be matched with any of the five women, the second man can be matched with one of the four remaining women, and so forth. By multiplying $\binom{10}{5}\binom{12}{5}5!$ by $5!$, you are saying that the order in which the couples are selected matters. $\endgroup$ Commented Jul 10, 2016 at 9:24

4 Answers 4

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I believe that the problem is that you're counting multiple times. If the couples were ordered (first couple, second couple, etc.) then it would be correct. In this case, though, you can simply go down the list of men and ask yourself "How many different women can I pair with this man?"

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    $\begingroup$ As a footnote, interestingly enough, this problem doesn't seem to explicitly forbid male/male or female/female couples, which would change the answer entirely. $\endgroup$
    – Mike
    Commented Jul 10, 2016 at 8:35
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It is $5!$ because you only need to arrange $5$ men (or $5$ women) while $5$ women (or $5$ men) are kept in a fixed order to pair them up. If you do $5!\cdot 5!$, that means you count duplicate pairs.

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Since you are arranging pairs of men and women, you can think of them as groups. After you have 5 men and 5 women chosen, you will have 5 groups that you need to arrange. This is why you only need to have $5!$ once in $\binom{10}{5}\binom{12}{5}5!$.

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After selecting 5 from men and 5 from women, we need to pair them. Lets assume like finding a pair for each man:
for the 1st guy -- can choose 1 from 5 women,
for the 2nd guy -- can choose 1 from 4 women,
$\vdots$
for the 5th guy -- can choose 1 from 1 woman
so 5!

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