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I can't figure out an algebraic proof for the following identity:

$$\sum_{i=0}^k{{n \choose i}{m \choose {k-i}}}= {{m+n}\choose k}$$

Combinatorical solution:

We can see that as choosing some from $n$ and the rest of $k$ from $m$, thus $k$ in total.

Or we could just choose $k$ from the union.

Martin Sleziak
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NightRa
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    For an algebraic proof (I don't know why you want one, to be honest), but consider the coefficient of $x^k$ in $(1+x)^n(1+x)^m = (1+x)^{n+m}$. – Andrew D Sep 11 '13 at 17:11
  • Its called the Chu-Vandermonde Identity, and there's already a post about it here: http://math.stackexchange.com/questions/219928/inductive-proof-for-vandermondes-identity – Salieri Nov 23 '13 at 13:48
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    @AndrewD: The algebraic proof does not depend on $n,m$ being natural numbers (but the binomial powers are now formal powers series); this is one reason why one might want an algebraic proof. – Marc van Leeuwen Nov 23 '13 at 14:57
  • Does this answer your question? Generalized Vandermonde's identity –  Jan 10 '22 at 05:16

3 Answers3

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For identities involving binomial coefficients sometimes "combinatorial proofs" and "algebraic proofs" are offered. Here we have two proofs for the Vandermonde Convolution:

  1. Algebraic proof: The sum of the LHS of the convolution formula equals the coefficient of $x^k$ on the LHS of the polynomial $$ (1+x)^n(1+x)^m=(1+x)^{m+n}. $$ The binomial coefficient on the RHS of the convolutions formula equals the coefficient of $x^k$ on the RHS of the polynomial equation.

  2. Combinatorial proof: Suppose that there are $n+m$ objects in a set, $n$ of them white and $m$ of them black. There are $\binom{n+m}{k}$ ways to choose $k$ elements in all. This is the RHS. The number of ways to choose $j$ white and $k-j$ black objects is the product $\binom{n}{j}\binom{m}{k-j}$, so the sum of all these objects, the LHS, must be the same total as on the RHS.

Dietrich Burde
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Using the binomial theorem consider $$\sum_{k=0}^{m+n}{m+n\choose k}x^k=(1+x)^{m+n}.$$ The right-hand side becomes $$(1+x)^m(1+x)^n=(\sum_{r=0}^m{m\choose r}x^r)(\sum_{s=0}^n{n\choose s}x^s).$$ The product of the two polynomials on the right-hand side can be rewritten as $$\sum_{k=0}^{m+n}\bigg{\{}\sum_{i=0}^k{n\choose i}{m\choose k-i}\bigg{\}}x^k.$$ Thus $$\sum_{i=0}^k{n\choose i}{m\choose k-i}={m+n\choose k}.$$

reyna
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1233dfv
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For another proof, see the First Proof of Theorem 3.29 in my Notes on the combinatorial fundamentals of algebra. (NB: It is Theorem 3.29 in the version of 10 January 2019. Theorem labels are subject to change, so search for "Chu-Vandermonde identity" if Theorem 3.29 is something different in your version.) This is not a particularly enlightening proof, but it has the advantage of requiring the least about $n$ and $m$: it works when $n$ and $m$ are elements of a commutative $\mathbb{Q}$-algebra. (Compare to the combinatorial proof, which requires $n$ and $m$ to be nonnegative integers. Compare also to the proof using the binomial formula, which works generally only if one has an algebraic way of proving $\left(1+x\right)^n\left(1+x\right)^m = \left(1+x\right)^{n+m}$ for non-integer $n$ and $m$.)

This has been linked from How do i prove that $\sum\limits_{r=0}^k \binom{m}{r}\binom{n}{k-r} = \binom{m+n}{k}$ , which has other proofs.

darij grinberg
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