Show that $\sum_{k=0}^n \binom{m-k}{n-k} = \binom{m+1}{n}$

Question

How can you show that:

$$\sum_{k=0}^n \binom{m-k}{n-k} = \binom{m+1}{n}$$

by using an index shift and the following relationship, which is assumed to be true:

$$\sum_{k=0}^n \binom{m+k}{k} = \binom{m+n+1}{n}$$

Answer 1

Hint: Use the following indentity (and induction on $m$) which is not hard to verify: $${n\choose k}+{n\choose k+1} = {n+1\choose k+1}$$

Answer 2

Replacing $m$ with $m-n$ gives $\sum_{k=0}^n\binom{m-n+k}{k}=\binom{m+1}{n}$. Changing the dummy variable viz. $k\mapsto n-k$ gives what you want.

Answer 3

Using the the symmetry of the binomial coefficients $\binom{n}{k}=\binom{n}{n-k}$ combined with $$ \sum_{i=0}^n\binom{i}{k}=\binom{n+1}{k+1} $$ with the understanding that $\binom{i}{k}=0$ for $i<k$ we wind up with

$$ \sum_{k=0}^n \binom{m-k}{n-k}= \sum_{k=0}^n \binom{m-k}{m-n}= \sum_{i=m-n}^m \binom{i}{m-n}= \sum_{i=0}^m \binom{i}{m-n} =\binom{m+1}{m-n+1}=\binom{m+1}{n} $$