Combinatorial proof for the identity $\binom{m + n}{r} = \binom{m}{0}\binom{n}{r} + \binom{m}{1}\binom{n}{r - 1} + \cdots + \binom{m}{r}\binom{n}{0}$

Question

Think of a set with $m+n$ elements as composed of two parts, one with $m$ elements and the other with $n$ elements. Give a combinatorial argument to show that

$\dbinom{m+n}{r}$ = $\dbinom{m}{0}$$\dbinom{n}{r}$ + $\dbinom{m}{1}$$\dbinom{n}{r-1}$ + $...$ + $\dbinom{m}{r}$$\dbinom{n}{0}$

where $n$ and $m$ are positive integers and $r$ is a integer that is less than or equal to both $n$ and $m$.

I don't have a idea to start thinking about that.

See this link or this answer. – Brian M. Scott Feb 07 '16 at 14:38 — Brian M. Scott, Feb 07 '16 at 14:38

score 4 · Accepted Answer · answered Feb 07 '16 at 15:06

The argument goes as follows:

L.H.S. represents the number of ways in which $r$ things can be chosen from $m+n$ things.

Now, the number of ways in which $r$ things can be chosen from $m+n$ things can also be arrived at in another procedure. We break the $m+n$ things into $m$ things and $n$ things. Hence choosing $r$ things from $m+n$ things can be done by choosing $0$ things from the $m$ things and $r$ things from the $n$ things; OR $1$ things from the $m$ things and $r$ things from the $n-1$ things; OR $2$ things from the $m$ things and $r$ things from the $n-2$ things; and so on.

Thus R.H.S. represents the sum of the number of ways in which these choices can be made.

So we can say that L.H.S.= R.H.S.

Combinatorial proof for the identity $\binom{m + n}{r} = \binom{m}{0}\binom{n}{r} + \binom{m}{1}\binom{n}{r - 1} + \cdots + \binom{m}{r}\binom{n}{0}$

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