Proving $\sum_{k=0}^\ell \binom{n}{k}\binom{m}{\ell-k}=\binom{n+m}{\ell}$ This is problem 4a of Spivak's Calculus, 4th Edition, Prologue Chapter 2:

Prove that $$\sum_{k=0}^\ell \binom{n}{k}\binom{m}{\ell-k}=\binom{n+m}{\ell}$$
Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$.

And this is the solution

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

It's frustrating when a solution uses terms like "clearly", and then you look at an equation and don't see it at all.

How is the coefficient of $x^l$ "clearly" the expression shown in the solution above?

 A: The lefthand side is of the form
$$(a_0 + a_1 x + \dots + a_n x^n)(b_0 + b_1 x + \dots + b_x x^m),$$
where $a_k = \binom{n}{k}$ and $b_j = \binom{m}{j}$. When we multiply everything out, when will terms of the form $c x^\ell$ appear?
Well, when a left term $a_k x^k$ meets a right term $b_j x^j$, we get a resulting term $a_k b_j x^{k + j}$. Thus, we are looking for pairs $(a_k x^k, b_j x^j)$ where $k + j = \ell$.
P.S. I agree with your sentiment that phrases like "clearly" are frustrating. You might like this article by Francis Su on the topic.
A: $\binom{n+m}{l}$ counts the ways of choosing $l$ items from a collection of $n+m$ items. Think your collection as having $n$ blue items and $m$ red ones. In order to choose $l$ items from your collection, you can first choose some blue ones, say $k$ of them, and then you can choose the rest $l-k$ from the red ones.
A: One formal derivation of the "clear" observation using a change in the index of summation $l = k + j$, or equivalently $j = l-k$.
\begin{align}
\sum_{k=0}^n \binom nk x^k \cdot  \sum_{j=0}^m \binom mj x^j &= 
\sum_{j=0}^m\sum_{k=0}^n \binom nk \binom mj x^{k + j}
\\ & = \sum_{l=0}^{m+n} \sum_{k=0}^l \binom nk \binom m{[l - k]}x^{k + [l-k]}
\\ & = \sum_{l=0}^{m+n} \sum_{k=0}^l \binom nk \binom m{l - k}x^{l}
 = \sum_{l=0}^{m+n} \left[\sum_{k=0}^l \binom m{l - k}\binom nk\right]x^{l}.
\end{align}
