I know that the following sum ($0 \le a \le b \le n$):
$$\sum_{k=a}^b{n \choose k}{k \choose r}$$
has a closed form when $a = 0$ and $b = n$:
$$\sum_{k=0}^n{n \choose k}{k \choose r} = 2^{n-r}{n \choose r}$$
(see e.g. Gould's Table of Combinatorial Identities vol. 2, page 4, (1.21)), but is it possible to simplify it in the other cases?