# Is my proof of $\binom{m+n}{r}=\binom{m}{0}\binom{n}{r} + \binom{m}{1}\binom{n}{r-1} + … + \binom{m}{r}\binom{n}{0}$ right?

As the title says, I was requested to prove

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

I was requested to do this using the following identity:

$$(1+x)^n(1+x)^m=(1+x)^{n+m}$$

Whenever I say identity I shall be meaning this identity; when a talk about the equation I shall be meaning the equation I was requested to prove.

I think I have found the proof, but I'm not sure if it's right (I'm fairly new to this subject), so I would like someone with more experience to help me deciding on this, and point my errors if any.

Firstly, I expand the binomials on the left-hand side of the identity. They are, according to the binomial theorem,

$$\sum_{r=0}^{n}\binom{n}{r}1^{n-r}x^{r} = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$$

and

$$\sum_{r=0}^m \binom{m}{r}1^{m-r}x^r=\binom{m}{0}+\binom{m}{1}x+\binom{m}{2}x^2+...+\binom{m}{m}x^m$$

Now that we expanded the factors of the left-hand side of our identity, we shall inspect the right-hand side of the equation and notice that it is the summation of the products of the expanded factors that are coefficient of $$x^n$$. For example, the first term of the right-hand side of the equation is $$\binom{m}{0}\binom{n}{r}$$. When we multiply $$(1+x)^n(1+x)^m$$, one of the coefficients of $$x^n$$ on the result is in deed the first element of the first factor multiplied by the last element of the second factor , $$\binom{m}{0}\binom{n}{n}x^n$$.

So now we know that the right-hand side of the equation is the coefficients of $$x^n$$ on the multiplication of $$(1+x)^n(1+x)^m$$. Because we know our identity is true, if we could prove that the $$x^n$$ coefficients of the right-hand of our identity is equal to the left-hand side of our equation, we would have proved our equation, since this would show both sides of our equation are just the $$x^n$$ coefficient of $$(1+x)^n(1+x)^m$$.

This can be shown expanding the right-hand side of our identity,

$$\sum_{r=0}^{n+m}\binom{n+m}{r}1^{n+m-r}x^r=\binom{n+m}{0}+\binom{n+m}{1}+...+\binom{n+m}{n}x^n+...+\binom{n+m}{n+m}x^{n+r}$$

As we can see, $$\binom{n+m}{n}$$ is the coefficient of $$x^n$$ too (just written differently). We know this is right because our identity is true. Let's call the $$x^n$$ coefficient $$C$$. If $$C=\binom{m}{0}\binom{n}{r} + \binom{m}{1}\binom{n}{r-1} + ... +\binom{m}{r}\binom{n}{0}$$ and $$C=\binom{m+n}{r}$$, then we proved our equation:

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

Is this proof right?

It sounds right, but I would write it like$$\sum_{k=0}^m\sum_{l=0}^n\binom{m}{k}\binom{n}{l}x^{k+l}=(1+x)^m(1+x)^n=(1+x)^{m+n}=\sum_{r=0}^{m+n}\binom{m+n}{r}x^r$$so that$$\binom{m+n}{r}x^r=\sum_{i=0}^r\binom{m}{i}\binom{n}{r-i}x^r.$$Then letting $$x=1$$ yields the desired expression.