Combinatorial proof of $\sum\limits_{i=0}^{r} ({m \choose i}) = {{m + r}\choose m}$ I'm having trouble generating a combinatorial proof for the following equality:
$$\sum_{i=0}^{r} ({m \choose i}) = {{m + r}\choose m}.$$ That is, with no algebraic manipulation, what is an "example" that would sufficiently illustrate what these expressions are doing? 
Here is my attempt at an answer: 
RHS: we have m distinguishable items of item type A, and $r$ distinguishable items of item type B. The number of ways to choose $m$ items from this combination of item type A and item type B is ${m + r} \choose m$.
LHS: there are $r + 1$ disjoint cases for which $r$ indistinguishable items are being categorized into $m$ distinguishable categories. And here is where I have trouble conceptualizing what the disjoint cases are.
Would really appreciate some help understanding conceptually why LHS equals RHS! 
 A: We can prove this by using the following equation: $\binom{m}{n}=\binom{m-1}{n}+\binom{m-1}{n-1}$
$\sum_{i=0}^r\left(\binom mi\right)=\sum_{i=0}^r\binom{m+i-1}{i}=\binom{m-1}{0}+\sum_{i=1}^r\binom{m+i-1}{i}=\binom{m}{0}+\sum_{i=1}^r\binom{m+i-1}{i}=\binom{m}{0}+\binom{m}{1}+\sum_{i=2}^r\binom{m+i-1}{i}=\binom{m+1}{1}+\sum_{i=2}^r\binom{m+i-1}{i}=\binom{m+2}{2}+\sum_{i=3}^r\binom{m+i-1}{i}=\cdot\cdot\cdot=\binom{m+r}{r}=\binom{m+r}{m}$
we prove the equality by applying $\binom{m}{n}=\binom{m-1}{n}+\binom{m-1}{n-1}$ 
repeatedly.
A: Here's a quick combinatorial proof, inspired by wikipedia's article on the hockey stick identity.
A committee of size $m$ from $m+r$ people can be formed in
$$\binom{m+r}{m}$$
ways. Now pass out the numbers $0,1,2,3,...,r$ to $r+1$ of the $m+r$ people. Now here are your $r+1$ disjoint cases. In case $i$ with $0 \leq i \leq r$, person $i$ is on the committee and persons $0,1,2,...i-1$ are not on the committee. This can be done in 
$$\binom{m+r-i-1}{m-1}=\binom{m+r-i-1}{r-i}$$
ways. Now we can write
$$\binom{m+r}{m}=\sum_{i=0}^r\binom{m+r-i-1}{r-i}=\sum_{i=0}^r\binom{m+i-1}{i}=\sum_{i=0}^r \left(\binom{m}{i}\right)$$
The only real bit of manipulation comes here in the form of
$$\left(\binom mi \right) = \binom{m+i-1}{i}$$
but this can be shown combinatorially as well through stars and bars.
