TITLE: Spivak's Calculus - Exercise 4.a of 2nd chapter
QUESTION [9 upvotes]: 4 . (a) Prove that
  $$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}.$$
  Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$.

I'm having a hard time trying to solve the problem above. I've done all of the previous exercises of the 2nd chapter with little difficulty, so far. I think I might be missing a trivial point somewhere.

The answer I got from the Answer Book, and is not very helpful either... :(

 4. (a) Since $(1+x)^n(1+x)^m = (1+x)^{n+m}$
 we have
 $$\sum_{k=0}^n \binom{n}{k}x^k\cdot\sum_{j=0}^m \binom{m}{j}x^j\cdot=\sum_{l=0}^{n+m} \binom{n+m}{l}x^l$$
 But the coefficient of $x^l$ on the left is clearly $$\sum_{k=0}^l\binom{n}{k}\binom{m}{l-k}.$$
 One term of the sum occurring for each pair $k$, $j = l-k$.  

I couldn't get the last part of the answer:
why is it that the "coefficient of $x^l$ on the left is clearly $\sum_{k=0}^l\binom{n}{k}\binom{m}{l-k}$."?

REPLY [11 votes]: Multiplication of formal power series is performed by collecting the terms with the same powers of $x$:
$$
\begin{align}
\left(\sum_{k=0}^\infty a_kx^k\right)\left(\sum_{k=0}^\infty b_kx^k\right)
&=\sum_{k=0}^\infty\left(\sum_{j=0}^k a_j\color{#C00000}{x^j}b_{k-j}\color{#C00000}{x^{k-j}}\right)\\
&=\sum_{k=0}^\infty\left(\sum_{j=0}^k a_jb_{k-j}\right)\color{#C00000}{x^k}\tag{1}
\end{align}
$$
Note that the subscripts in the inner sum add up to $k$, the power of $x$ in the outer sum.
Apply $(1)$ to the product of
$$
(1+x)^m=\sum_{k=0}^\infty\binom{m}{k}x^k\tag{2}
$$
and
$$
(1+x)^n=\sum_{k=0}^\infty\binom{n}{k}x^k\tag{3}
$$
which is
$$
(1+x)^{m+n}=\sum_{k=0}^\infty\binom{m+n}{k}x^k\tag{4}
$$
I extended the indices in the sums to $\infty$ since for $k>n$, $\binom{n}{k}=0$.
For the product of $(2)$ and $(3)$, we get
$$
(1+x)^m(1+x)^n=\sum_{k=0}^\infty\left(\sum_{j=0}^k \binom{m}{j}\binom{n}{k-j}\right)x^k\tag{5}
$$
Comparing the coefficients of $x^k$ in $(4)$ and $(5)$ yields
$$
\binom{m+n}{k}=\sum_{j=0}^k \binom{m}{j}\binom{n}{k-j}\tag{6}
$$
as desired.