Number of multisets on $[2m]$ which satisfy certain conditions. I am trying to find the number of $n$-element multisets on $[2m]=\left\{1, \ldots, 2m\right\}$ such that $m+1, \ldots, 2m$ appear an even number of times in the $n$-multiset.
I have tried several approaches, but the following seems to be the best approach so far...
Let $A=\left\{m+1, \ldots, 2m\right\}$ and let $\nu(i)$ be the number of times $i\in [2m]$ appears in the multiset. We're trying to find the number of multisets $\left\{1^{\nu(1)}, 2^{\nu(2)} \ldots, 2m^{\nu(2m)} \right\}$ such that $\sum_{i=1}^{2m}\nu(i)=n$ and $\nu(i)=2j_i$, for some $j_i\in \mathbb{N}$ where $i\in A$. This is equivalent to finding the number of weak compositions of $n$ with $2m$ parts, $\sum_{i=1}^{2m} x_i=n$, such that $x_i=2j_i$ for each $i\in A$. 
There are two cases:
$\textbf{Case 1:}$ $j_i=0$ for all $i\in A$. In this case, $\sum_{i=1}^{2m} x_i=\sum_{i=1}^{m} x_i=n$, which is a weak composition of $n$ into $m$ parts, of which there are $\binom{n+m-1}{m-1}$.
$\textbf{Case 1:}$ $j_i=0$ for all $i\in S$, where $S\in \binom{A}{k}$ (i.e., $S\subseteq A$ such that $|S|=k$) and $j_i\geq 1$ for all $i\in A\setminus S$. In this case,
$$\sum_{i=1}^{2m} x_i=\sum_{i=1}^{m} x_i+\sum_{i\in A\setminus S}x_i+\sum_{i\in S}x_i=\sum_{i=1}^{m} x_i+\sum_{i\in A\setminus S}x_i=n$$, which is a weak composition of $n$ with $2m-k$ parts such that $x_i=2j_i$ for all $i\in A\setminus S$.
Here is where I hit a brick wall! I don't see a simple way to figure out the number of weak compositions of $n$ with $2m-k$ parts such that $x_i=2j_i$ for all $i\in A\setminus S$. Your help is extremely appreciated!!
 A: You are looking for the coefficient of $x^n$ in $((1-x)(1-x^2))^{-m}$, so all you have to do is convolve the sequence $a_i=\binom{m-1+i}{m-1}$ with its aerated version. For example, if $m=4$, then write down: $$\begin{array}{rrrrrrrr}\\1&4&10&20&35&56&84&120&\ldots\\1&0&4&0&10&0&20&0&\ldots\end{array}$$
Then, cross-multiplying:
$$1\cdot1=1\\
1\cdot0+4\cdot1=4\\
1\cdot4+4\cdot0+10\cdot1=14\\
1\cdot0+4\cdot4+10\cdot0+20\cdot1=36\\
1\cdot10+4\cdot0+10\cdot4+20\cdot0+35\cdot1=85$$
So for example, there are $36$ three-element and $85$ four-element multisets on $[8]$ with the property you want.
A: I see that I'm over five years late, but nice problem. :) We present a bijective solution, which I feel is pretty hard to find without using generating functions first.
We claim that the answer is $\binom{m+n-1}n$ ($m$ multichoose $n$).
Suppose we are given an $n$-multiset $S$ with elements in $[2m]$ satisfying the given property. Define $T$ with elements in $[m]$ so that if $k \in [m]$ appears $a_k$ times in $S \cap [m]$ and $k+m$ appears $b_k$ times in $S \cap ([2m]-[m])$, $k$ appears $a_k+b_k$ times in $[m]$.
Conversely, suppose we are given an $n$-multiset $T$ with elements in $[m]$. We can retrieve the $n$-multiset $S$ with elements of $[2m]$ as follows: suppose there are $a_k$ instances of $k \in [m]$ in $T$. In $S$, leave $a_k \bmod 2$ of them as $k$, and $a_k - (a_k \bmod 2)$ to $k+m$.
It is clear that both of these are injective and that the resulting $T$ and $S$ actually satisfy the necessary conditions, so this is a bijection. Hence the number of possible $S$ equals the number of possible $T$, $m$ multichoose $n$.
@Andrew Woods: The generating function is actually
\begin{align*}
(1+x)^m (1+x^2+x^4+\cdots)^m &= (1+x)^m (1-x^2)^{-m} \\
&= \left(\sum_{k \ge 0} \binom mkx^k\right) \cdot \left(\sum_{j \ge 0} \binom{m+j-1}j x^{2j}\right) \\
&= \sum_{n \ge 0}\left[\sum_{k+2j = n} \binom mk \binom{m+j-1}j\right]x^n
\end{align*}
From here, I tried $(m,n) = (4,4),(5,4),(6,4)$ to get the desired formula, though I'm sure there's a combinatorial way to evaluate this.
