Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$

Question

To prove $$\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i},$$

I demonstrated that the equality is true for any $n,$ for $m = 0, 1,$ and for any $r < n + m$ simply by fixing $n$ and $r$ and inserting $0,1$ for $m.$ Then, I proceed to induct on $m$ (and on $m$ only).

---

I am not perfectly confident in my self, however, for I see two placeholders, $n$ and $m.$ Is this a case where double induction is needed (first on $m$ and then on $n$)?

---

Whether or not this proof requires double induction, may someone explain when double induction is needed?

---

Consider any fixed $n, r \geq 0$ and the following two cases (I know that only one case is needed to complete this inductive proof).

CASE 1

\begin{align} \binom{n + 0}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{0}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r}\binom{0}{0} \\ &= 0 + 0 + \cdots + \binom{n}{r} \\ &= \binom{n}{r} \end{align}

CASE 2

\begin{align} \binom{n + 1}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{1}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r-1}\binom{1}{r - (r-1)} + \binom{n}{r}\binom{1}{r - r} \\ &= 0 + 0 + \cdots + \binom{n}{r-1} + \binom{n}{r} \\ &= \binom{n}{r-1} + \binom{n}{r} \end{align}

INDUCTION

Suppose it is true for $m \leq k.$ Now, consider $$\binom{n + (k + 1)}{r}.$$ It follows from Pascal's Identity that

$$\binom{n + (k+1)}{r} = \binom{n + k}{r} + \binom{n + k}{r-1}$$

And,

\begin{align} \binom{n + k}{r} + \binom{n + k}{r-1} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\bigg[\binom{k}{r - i} + \binom{k}{r - 1 - i}\bigg] \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k+1}{r-i} \\ &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k+1}{r-i} \end{align}

Hence, the equality holds for $m = k + 1.$ Given that the equality holds for $m = 0, 1,$ and that if equality holds for $m = k,$ it then holds for $m = k + 1,$ it follows that the equality holds $\forall m \in \mathbb{N}.$

Answer 1

For a short reply, your induction proof has tiny problem, in that $r$ can take value $n+k$ for the $m = k+1$ induction step. So when you use induction hypothesis to get ${{n+k}\choose{r}} = \sum_{i=1}^r {n\choose i}{k\choose {r-i}}$, you can actually use it only for $r<n+k$. This is not big problem though as the $r = n+k$ case is trivial.

For the double-induction, I don't think it's necessary here. The reason is that you are actually fixing an arbitrary $n$ first, and then do induction proof. So the induction proof is within the context of the fixed $n$.

Answer 2

I think the formula is wrong. If you have $n$ white balls and $m$ red balls then, to choose $r$ balls from then you have to choose $i$ white balls and $r-i$ red balls for some $i$ between $0$ and $\min \{r,n\}$. Hence the sum should be over $0\leq r \leq \min \{r,n\}$. However, if you define $\binom {n} {i}$ to be $0$ when $i >n$ then the formula is correct.

Answer 3

You can think of a combinatoric proof as follows.

Suppose we have $n$ men and $m$ women, and we wish to form a committee of $r$ people. We can count in two different ways.

Case 1 $\binom{n+m}{r}$ is the number of r-subsets of a set with n+m elements.

Case 2 First we pick $i$ males. That leaves $r-i$ female to choose. So given $0\le i \le r$ We have $\binom{n}{i}\binom{m}{r-i}$ such committees. If we let $i$ range, we add up all these committees, so we get $\displaystyle\sum_{i=0}^r \binom{n}{i}\binom{m}{r-i}$ .

Since both cases count the same number, they must be equal.