Generating Function of 'Vandermonde Identity' Like Coefficients?

Question

I wish to simplify the following generating function: $$ f(x) = \sum_{j = 0}^k\binom{k}{j}\binom{n - k}{m - j}x^j. $$ Note that Vandermonde's identity tells us that: $$ f(1) = \sum_{j = 0}^k\binom{k}{j}\binom{n - k}{m - j} = \binom{n}{m}. $$ Hence, the title! Any help will be appreciated.

Answer 1

Write $f(x)$ as $f_m(x)$ to indicate the dependence on $m$. By binomial theorem, we see that $f_m(x)$ is the coefficient of $y^m$ in the product $$ (1+xy)^k (1+y)^{n-k}.$$

Hence, we are able to write a generating function $$ \sum_{m=0}^{\infty} f_m(x) y^m = (1+xy)^k(1+y)^{n-k}. $$

To find $f_m(x)$ from this, take $m$-th derivative of the RHS and put $y=0$. Then we have $$ m! f_m(x) = \left(\frac {\partial}{\partial y}\right)^m (1+xy)^k(1+y)^{n-k} \Bigg\vert_{y=0}. $$