How find this sum $\sum_{k=0}^{p}t^k\binom{n}{k}\binom{m}{p-k}$

Question

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

Answer 1

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.

Answer 2

Hint: this equation represents coefficient of $x^p$ in $$(1+tx)^n×(1+x)^m$$

Answer 3

There is a closed form but I am almost sure you will not enjoy it. Using a CAS,

$$\sum_{k=0}^{p}t^k\binom{n}{k}\binom{m}{p-k}=\binom{m}{p} \, _2F_1(-n,-p;m-p+1;t)$$