How find this sum $\sum_{k=0}^{p}t^k\binom{n}{k}\binom{m}{p-k}$ Find the closed form
$$\sum_{k=0}^{p}t^k\binom{n}{k}\binom{m}{p-k}$$
since
$$\binom{n}{k}\binom{m}{p-k}=\dfrac{n!}{(n-k)!k!}\cdot\dfrac{m!}{(p-k)!(m-p+k)!}$$
then I can't 

Let's find the generating function $F(z):=\sum_p a_pz^p$, where $a_p$ is your sum.
Notice that your sum is a convolution of $t^k\binom{n}{k}$ and $\binom{m}{k}$.
Therefore $$\begin{align}F(z)&=\left(\sum_k t^k\binom{n}{k}z^k\right)\left(\sum_k\binom{m}{k}z^k\right)\\&=(1+tz)^n(1+z)^m\end{align}$$
Notice that for $m=0$, for example, the sum is $\sum_{k=0}^{p}\binom{n}{k}t^k$, which for $p=0,1,...,n$ are the truncations of $(1+t)^n$. These don't have a closed(er) form.
