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 :
- I choose 5 women from a pool of 10 in $^{10}C_2$ ways.
- 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
$\endgroup$ 63 Answers
$\begingroup$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?"
$\endgroup$ 1 $\begingroup$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
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
$\endgroup$ $\begingroup$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! * 5!, that means you count duplicate pairs.
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