Combinatorial Proof of $\binom{i+j}{j}S(n, i+j) = \sum_{k=0}^{n}\binom{n}{k}S(k,i)S(n-k,j)$ I would like to prove the following statements:
$\binom{i+j}{j}S(n, i+j) = \sum_{k=0}^{n}\binom{n}{k}S(k,i)S(n-k,j)$
combinatorially.
(where $\;S(i,j),\;i,j\in \Bbb Z\;$ denotes the Stirling number of the second kind, put labelled $j$ entities into $i$ unlabeled classifications) 

I looked over related Wikipedia pages and checked over some recurrence identities of Stirling numbers of the second kind, however, there's no recurrences that decompose $S$ into two different $S$s.
Moreover, I need to understand combinatoric understanding behind this identity which is the final goal of mine.
Any advice where to start from? (maybe Pascal's triangle?)

HINT: Show that this is the number of ways to put $n$ labelled objects into $i$ red boxes and $j$ blue boxes.  For the full combinatorial argument, hover below:

 We can count these in two ways: first, just put $n$ objects into $i + j$ uncolored boxes, and then choose which boxes will be colored blue.  This corresponds to $S(n,i+j)\cdot\binom{i+j}{j}$  The second way is to start with $i$ and $j$ red boxes, and let $k$ be the number put into red boxes.  So we have to sum over $k$, choose which $k$ elements go in red boxes, and then put the $k$ elements in the $i$ red boxes and $n-k$ elements in the $j$ blue boxes.  This is the right hand side.

