The following admits of many (easy) proofs, but I am seeing no purely "bijective" argument:


$$
\sum\_{j=n}^N \binom{j}{n} = \binom{N+1}{n+1}.
$$


Any ideas?



Combinatorial proof of identity
Santa Claus has $N+1$ reindeer whose noses are of varying redness. Every year, Santa needs $n+1$ reindeer to pull his sleigh. The reddest-nosed reindeer always leads the sleigh.


The way Santa chooses the reindeer is as follows. First, Santa chooses one reindeer to lead the sleigh; call it the $(j+1)$th reindeer. Then out of the $j$ reindeer whose noses are less red, Santa chooses $n$ more reindeer to complete the team.


One year, Santa learns (on MistletOverflow) that he can simply choose $n+1$ reindeer, and then see which of them has the reddest nose.


