Give a bijective proof to show that the number of ways presesnts can be distributed? I am asked the following: 
We want to distribute $n$ presents to $k$ children. We are told that we are supposed to give child $1$ $a_{1}$ presents, child $2$ $a_{2}$ presents,$\ldots$, child $k$ $a_{k}$ presents. 
(Assume that $a_{1}+\ldots+a_{k}=n$).
Recall that $\binom{n}{a_{1},\ldots,a_{k}}$ is defined to be the sequences of length $n$ with $a_{i} i$'s, for $1\leq i\leq k$.
Give a bijective proof to show that the number of ways these presents can be distributed is also counted by $\binom{n}{a_{1},\ldots,a_{k}}$.

This is from my first homework assignment for the quarter, however we haven't talked too much about bijective proofs.
I'm unsure what we should consider for a bijection, more specifically the mapping.
Do I want to consider the set of children, $C=\{1,\ldots ,k\}$, and the set of  presents, denoted by $P=\{a_{1}, \ldots ,a_{k}\}$; take the power set of both $P$ and $C$, then define a function $\varphi:2^{C}\rightarrow2^{P}$? What should the mapping be? 
Any push is GREATLY appreciated.
 A: You could count the number of ways to distribute presents directly and show that this is exactly the multinomial combination. But if you are looking for a "bijective" proof, you, as SE318 suggests, want to create a bijection between the set of sequences of length $n$ with $a_i$ $i$'s and the set of ways to distribute the presents. You're initial inclination is interesting, but I don't think it will get you anywhere because you should be interested in a bijection between the previously mentioned sets. Most of the time with proofs like these, you don't want to rigorously define a function and prove that it's a bijection, but give a "transformation" that shows that these problems are the same and so have the same solution. Let me show you:
Let $A$ be the number of sequences of length $n$ with $a_i$ $i$'s and let $B$ be the number of ways to distribute these presents.
Fix the order of the presents, $p_i, \ldots, p_n$. Given a sequence of length $n$ with $a_i$ $i$'s, let an $i$ in the $j$th spot of the sequence mean that we give present $p_j$ to kid $i$. This is valid for any sequence (you should check this), so $A \geq B$. 
Similarly, given a way to distribute present you can derive a sequence just as above. This is valid for any present distribution (check), so $A \leq B$. Thus we've created a bijection between the two problems, but haven't talked about formal functions.
