Finite sum of combinations Suppose $a$ and $b$ are two positive integers with $a \leq b$. Do we have the following?
$$\sum_{k=0}^{a}\binom{a}{k}\binom{b}{b-k} = \binom{a+b}{a}$$
This is very intuitive to me: we want to choose $a$ items from a set of $a+b$ items. We can choose $k \leq a$ items from the first $a$ items and $b-k$ items from the last $b$ items.

Note that
$$ (1+x)^{a+b}=(1+x)^a(1+x)^b. $$
By using
$$ (1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k $$
one has
$$ (1+x)^{a+b}=\sum_{k=0}^n\binom{a+b}{k}x^k \tag1$$
and
$$ (1+x)^a(1+x)^b=\sum_{k=0}^a\binom{a}{k}x^k\sum_{k=0}^b\binom{b}{k}x^k=\sum_{j=0}^a\sum_{k=0}^b\binom{a}{j}\binom{b}{k}x^{j+k}. \tag2 $$
From (1), the coefficient of $x^a$ is
$$\binom{a+b}{a} $$
and from (2), the coefficient of $x^a$ is
$$ \sum_{j=0}^a\binom{a}{j}\binom{b}{a-j}. $$
So
$$ \sum_{j=0}^a\binom{a}{j}\binom{b}{a-j}=\binom{a+b}{a}. $$
