Simplification of the sum $\sum_{k=0}^M x^k\binom{M}{k}\binom{r}{k}?$

Question

For any $r\in\mathbb{R}$ and $k\in\mathbb{N}$ let $$\binom{r}{k}=\frac{r(r-1)(r-2)...(r-k+1)}{k!}$$ be a generalized binomial coefficient.

For $k, M\in\mathbb{N}$ and $r,x\in\mathbb{R}$ is there a way to calculate/simplify the expression $$\sum_{k=0}^M x^k\binom{M}{k}\binom{r}{k}?$$

Answer 1

$$\sum_{k=0}^M \binom{M}{k}\binom{r}{k}x^k=\, _2F_1(-M,-r;1;x)$$ where appears the gaussian hypergeometric function