Here is a "colourful" description of what I would like to count. Suppose you have one of those tables you see in a casino. I think they are for roulette, with $m$ squares, each of them with a number inside them. You have $c_1$ chips of type 1, $c_2$ chips of type $2,\dots,c_n$ chips of type $n$. You have to place the chips on top of the squares such that each number has at least $1$ chip over it but no square has two or more chips of the same colour on top of it. In how many ways can we do this?
1 Answer
The answer is the coefficient of $x_1^{c_1}\cdots x_n^{c_n}$ in $[(1+x_1)(1+x_2)\cdots(1+x_n)-1]^m$. Here $(1+x_1)\cdots (1+x_n) -1$ corresponds to the chips placed on each square; zero or one of each color, but at least one chip. We can expand by the binomial theorem to get that this coefficient is $$\sum_{k=0}^m (-1)^{m-k}\binom mk \binom k{c_1}\cdots \binom k{c_n}.$$ This formula can also be proved directly by inclusion-exclusion: $\binom mk \binom k{c_1}\cdots \binom k{c_n}$ is the number of ways to choose $k$ of the $m$ squares and place all of the chips on these squares, with at most one chip of each color on each square (but with some squares possibly having no chips).
-
$\begingroup$ Can I use this for a question in math.se? $\endgroup$– GorkaCommented Jun 2, 2014 at 15:19
-
$\begingroup$ I don't understand what you are asking. $\endgroup$ Commented Jun 3, 2014 at 0:01
-
$\begingroup$ can I use your solution for a different problem in another website ? the url of that website is math.stackexchange.com but most people just call it math.se $\endgroup$– GorkaCommented Jun 3, 2014 at 0:04
-
$\begingroup$ As far as I'm concerned you can. $\endgroup$ Commented Jun 3, 2014 at 4:21