Permutations/combinations question:
Number of ways 24 students of 12 boys and 12 girls, how many ways can be separated into groups of 3 if all groups have to be mixed gender?
I thought I had the answer when I did this:
- Take 8 boys, and put them each into one group. This can be done in 8! ways.
- Take 8 girls, and put them each into one group. Once
again, 8! ways. - There are 4 girls and 4 girls left, and you can put either a boy or a girl in the last position, so that’s 8!x8!x8!
But, This can’t be right because it doesn’t the times when, for example, two of the boys from the group of 8 are in the same group. Can anyone help me out?
Best Answer
You're idea is right, but it needs some refinement. First you have to select the $8$ boys in the first step and similarly the $8$ girls in the second one. Moreover for the last step we have $8!$ ways to do it. Indeed the first boy can be placed in $8$ groups, the next one in $7$ and so on.
Finally to avoid double counting you have to divide by $(2!)^8$. This is because each group has $2$ boys (or girls) and each of it can be used in the first step of creating the very same group.
Additionally you want to divide by $8!$, as I assume the group are undistinguishable. I believe that having placing boy $1,2$ and girl $1$ in the first group is same as placing them in the second one. Thus the final answer is:
$$\frac{\binom{12}{8} \cdot 8! \cdot \binom{12}{8} \cdot 8! \cdot 8!}{(2!)^8 \cdot 8!}$$