I have a combinatorial problem to solve, since I haven't done this for a while it would be great to get some feedback as to whether my argument is correct. Here it goes:
Suppose we have six slots and six balls to allocate (one per slot). The balls are of three different colors (so two per each color), say black (b), white (w), and red (r).
How many ways are there to allocate these balls into the slots, given that the ordering of the colors does not matter, so that for example ssbbww = bbwwss, sbswwb = wswbbs and so forth.
I argued as follows: \begin{equation} \text{There are } \binom{6}{2} \text{ ways to drop two balls of one color } \end{equation} and \begin{equation} \binom{4}{2} \text{ ways to choose the slots for the next color. } \end{equation} Dropping the remaining two balls has no freedom as only two slots are left. In order to account for the fact that the ordering of the colors (i.e. which color I choose first, second, third) does not matter I then have to divide by $3\cdot2$ as these are the number of possible ways to choose amongst three options succesively. So I get to the answer \begin{equation} \binom{6}{2}\binom{4}{2}\frac{1}{3\cdot2} = 15 \end{equation} Is that reasoning correct ? Many thanks !