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 

There is a closed form but I am almost sure you will not enjoy it. Using a CAS, 
$$\sum_{k=0}^{p}t^k\binom{n}{k}\binom{m}{p-k}=\binom{m}{p} \, _2F_1(-n,-p;m-p+1;t)$$
