Proving $\sum_{k=0}^\ell \binom{n}{k}\binom{m}{\ell-k}=\binom{n+m}{\ell}$

Question

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?

Have you tried considering an example? For instance, $m = 2$ and $n = 3$. — Ben Grossmann, Feb 01 '21 at 19:23
I agree that the use of "clearly" here is unwarranted and frustrating — Ben Grossmann, Feb 01 '21 at 19:25
This is Vandermonde's identity, asked many times (e.g. here). — metamorphy, Feb 01 '21 at 19:38
It’s just the usual product of two polynomials. If you write them out instead of using a summation, it should be fairly clear. Or you could observe that the coefficient of $x^\ell$ on the left is the sum of all terms $\binom{n}k\binom{m}j$ such that $k+m=\ell$, and since that means that $m=\ell-k$, that gives you the desired result. — Brian M. Scott, Feb 01 '21 at 20:56
See explanation here: https://math.stackexchange.com/a/3951400/754927 — Ben, Feb 01 '21 at 21:38

score 5 · Accepted Answer · answered Feb 01 '21 at 19:32

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.

posilon · Answer 2 · 2021-02-01T20:50:12.050

2

$\binom{n+m}{l}$ 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.

edited Feb 01 '21 at 20:50

answered Feb 01 '21 at 19:36

posilon

2,253

Ben Grossmann · Answer 3 · 2021-02-01T23:06:10.747

1

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}

edited Feb 01 '21 at 23:06

answered Feb 01 '21 at 19:39

Ben Grossmann

225,327

Are there errors in the summation indices above? For example, doesn't j go from 0 to m? Once you change the index to l = k + j, doesn't l go from i to m+n? – xoux Feb 01 '21 at 22:55
@evianpring I indeed made mistakes with the indices. See my latest edit. Note that the double summation is meant to denote a nested sum, so it doesn't make sense for the limits of $l$ to depend on the "internal" index $k$. – Ben Grossmann Feb 01 '21 at 23:08
Ok, I need to learn a bit more about when and how to change summation indices. – xoux Feb 02 '21 at 00:11

Proving $\sum_{k=0}^\ell \binom{n}{k}\binom{m}{\ell-k}=\binom{n+m}{\ell}$

3 Answers3