# Prove combinatorial identity using inclusion/exclusion principle

The identity is:

$$\sum_{k=0}^{m} (-1)^{k} {{n} \choose {k}}{{n-k}\choose{m-k}} = 0$$

I'm not even sure where to begin. Does anyone have any suggestions?

Let $A$ be an $n$-element set, and assume that $0<m\le n$; we’ll count the $m$-element subsets of $A$ that don’t contain any element of $A$ in two different ways. First, of course, it’s obvious that there are no such subsets. (This is why we have to take $m>0$: if $m=0$ there actually is one such subset, $\varnothing$.)

Now we’ll use inclusion-exclusion. As a first approximation there are $\binom{n}m$ $m$-subsets of $A$. Let $a\in A$ be arbitrary; there are $\binom{n-1}{m-1}$ $(m-1)$-subsets of $A\setminus\{a\}$, so there are $\binom{n-1}{m-1}$ $m$-subsets of $A$ that contain $a$. Since we want to count only those $m$-subsets of $A$ that don’t contain $a$, we need to subtract $\binom{n-1}{m-1}$. Moreover, we need to do this for each of the $\binom{n}1$ elements of $A$, so a second approximation to the desired number is

$$\binom{n}m-\binom{n}1\binom{n-1}{m-1}\;.\tag{1}$$

Now let $a$ and $b$ be distinct elements of $A$; every $m$-subset of $A$ that contains both $a$ and $b$ has been subtracted twice in $(1)$, once for $a$ and once for $b$, so we need to add these sets back in. There are $\binom{n-2}{m-2}$ ways to choose $m-2$ more elements of $A$ to expand $\{a,b\}$ to an $m$-subset of $A$, so there are $\binom{n-2}{m-2}$ $m$-subsets of $A$ that contain both $a$ and $b$. And there are $\binom{n}2$ possible pairs $\{a,b\}$ of distinct elements of $A$, so we have to add back $\binom{n-2}{m-2}$ for each of them, getting as a third approximation

$$\binom{n}m-\binom{n}1\binom{n-1}{m-1}+\binom{n}2\binom{n-2}{m-2}\;.$$

I expect that by now the full inclusion-exclusion argument is pretty clear.

By the way, there is also a combinatorial argument that doesn’t use inclusion-exclusion. Note first that

$$\binom{n}k\binom{n-k}{m-k}$$

is the number of ways to choose $k$ elements of $A$ and paint them red, and then choose $m-k$ of the remaining $n-k$ elements and paint them blue. It follows that

$$\sum_{k=0}^m\binom{n}k\binom{n-k}{m-k}\tag{2}$$

is simply the number of ways to choose $m$ elements of $A$, paint any subset of them red, and paint the remainder of the chosen elements blue. Let

$$\mathscr{P}=\{\langle S,R\rangle:S\text{ is an }m\text{-subset of }A\text{ and }R\subseteq S\}\;;$$

if you think of $R$ as the subset of $S$ that’s painted red (so that $S\setminus R$ is painted blue), it’s not hard to see that the sum in $(2)$ is simply $|\mathscr{P}|$. The sum

$$\sum_{k=0}^m(-1)^k\binom{n}k\binom{n-k}{m-k}\tag{3}$$

in a sense also counts $\mathscr{P}$, but it counts each $\langle S,R\rangle\in\mathscr{P}$ with a plus sign if $|R|$ is even and with a minus sign if $|R|$ is odd.

Let $S$ be any $m$-subset of $A$. Since $m>0$, $S$ has $2^{m-1}$ subsets of odd cardinality and $2^{m-1}$ subsets of even cardinality. Thus,

$$\sum_{R\subseteq S}(-1)^{|R|}=2^{m-1}-2^{m-1}=0\;,$$

and we see that $\{\langle S,R\rangle:R\subseteq S\}$ contributes $0$ to the sum in $(3)$. This is true for every $m$-subset $S$ of $A$, so the sum in $(3)$ must be $0$.

First see that the identity holds only when $m>0$. For $m=0$, L.H.S equates to $1$.

Proof by Double Counting

Q: Count the number of ways of selecting m red balls from a set of n blue balls and r red balls. Balls are distinct and numbered 1 to n and 1 to r respectively.

Answer 1: $r \choose m$

Let $A_i$ is the number of selections of size m from the set of n blue and r red balls containing i-th blue ball.
Let Z denote the set of n blue balls and r red balls. Then, $A_\phi = {{n+r}\choose m}$ and $A_{I:|I|=k} = {{n+r-k}\choose{m-k}}$ and by the Principle of Exclusion-Inclusion,
The number of selections of size m containing only red balls from the set of n blue and r red balls = $$\sum_{k=0}^m {(-1)^k{n \choose k}{{n+r-k}\choose {m-k}}}$$
Now set $r=0$ to get,
$$\sum_{k=0}^m {(-1)^k{n \choose k}{{n-k}\choose {m-k}}} = {0 \choose m} = 0$$ Since $m>0$.[Proved]