Say you have $100$ socks, where $50$ are black, $30$ are white, $20$ are blue. You want to distribute them to $10$ different people such that none of them end up with an odd number of any colour, but there is no requirement that all $10$ people get socks. We are assuming the socks are identical other than colour and the people are all distinguishable. My solution is:
There are $25$ pairs of black socks, $15$ white, $10$ blue. There are
$\binom{10+10-1}{10}=\binom{19}{10}$ ways to distribute the blue socks,
$\binom{25+10-1}{25}=\binom{34}{25}$ ways to distribute the black socks, and
$\binom{15+10-1}{15}=\binom{24}{15}$ ways to distribute the white socks.
So there are $\binom{19}{10}\binom{34}{25}\binom{24}{15}$ total ways to distribute the $100$ socks such that nobody gets an odd number of any of the socks.
Would anybody be able to check my solution for this?
I was looking into the Generating Function approach to this question. The 25 black pairs of socks could be represented by; $$(1+x+x^2+x^3+x^4+...+x^{25})$$ and the ten boxes by raising this to the power of 10; $$(1+x+x^2+x^3+x^4+...+x^{25})^{10}$$ So, essentially, the answer could be obtained from expanding the brackets of; $$(1+x+x^2+...+x^{25})^{10} \times(1+y+y^2+...+y^{15})^{10} \times(1+z+z^2+...+z^{10})^{10}$$ and looking at the coefficient of $x^{25}y^{15}z^{10}$ where $x,y,z$ represent black, white and blue pairs of socks respectively. This looks heavy going, although each part is a geometric progression; $$\bigr(\frac{(1-x^{25})}{(1-x)}\bigr)^{10} \times \bigr(\frac{(1-y^{15})}{(1-y)}\bigr)^{10} \times\bigr(\frac{(1-z^{10})}{(1-z)}\bigr)^{10}$$ Keeping in mind that we're only after the coefficient of $x^{25}y^{15}z^{10}$ it might be possible to extract it$\dots$ Anyone want to take it on from here ?