Representatives of three research institutes should form a commission of 9 researchers. How many ways can this committee be formed such that no institute should have an absolute majority in the group?
My partial solution:
I will use the exponential generating function because the researchers of the three institutes are different. The maximum number of researchers at an institute is 8, so the generating function will be given by:
$$f(x) = \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + \frac{x^7}{7!} + \frac{x^8}{8!}\right)^3$$
Is it correct here? How do I proceed?
You should not use an exponential generating function here. You would use this if your were choosing an ordered lists of nine researchers. Here, we are just choosing how many representative appear from each institution, so ordinary generating functions are applicable.
The number of representatives can be anywhere between $0$ and $4$, because if there were $5$ or more, there would be an absolute majority. Therefore, the generating function for a single institution is $(1+x+\dots+x^4)=(1-x^5)/(1-x)$. Since there are three institutions, the generating function is the cube of that, $$ (1-x^5)^3(1-x)^{-3}\tag{$*$} $$ Finally, we want the coefficient of $x^9$ in the above expression. This is a product of two nice functions, $(1-x^5)^3$ and $(1-x)^{-3}$ so the coefficient of $x^9$ in the product is the convolution of these two sequences. Namely, since $$ (1-x^5)^3= 1-3x^5+3x^{10}-x^{15},\quad \text{and}\\ (1-x)^{-3}=\sum_{k\ge 0}\binom{k+2}{2}x^k $$ it follows that the coefficient of $x^9$ in $(*)$ is $$ \binom{9+2}{2}-3\binom{4+2}2 $$