Words of length $n$ consisting of $\{A,B,C,D\}$, where $\#A=\#B$ and $\#C=\#D$. For given length $n$, calculate number of words consisting of letters $\{A,B,C,D\}$, where $\#A=\#B$ and $\#C=\#D$.
I know that for odd $n$ the answer is $0$.
Otherwhise, I think that the result is equal to
$\sum_{j=0}^{\frac{n}{2}} \binom{n}{j} \binom{n-j}{j} \binom{n-2j}{n/2 -j}$ (we choose positions for $A$, then for $B$, then for $C$, and the remaining positions for $D$), but I am curious whether we can calculate the result in more simple way.
 A: Let $n=2k$
Imagine this as a 2-dimensional random walk.  Call "B" the positive $x$ direction, call "D" the positive $y$ direction, meanwhile call $A$ the negative $x$ and $C$ the negative $y$ directions respectively.  Now... ignoring the exact letters, you know that exactly half of the positions will be in some sort of "positive direction" whether that was positive $x$ or positive $y$.  There are $\binom{2k}{k}$ ways to make such choices.
Now, rotate the image 90 degrees.  We observe the same thing, exactly half must be pointed in a (currently) positive direction.  Choosing half to do so will then uniquely determine what exact random walk, and as such what string of letters, it was that we are dealing with.  Reversing the process is also possible allowing you to prove that this is a bijection.  The final answer then:
$$\binom{2k}{k}^2$$
A: If $\#A=\#B\in [0,n/2] $ is given, then $\#C=\#D = (n-2\#A)/2 = n/2-\#A$.  The number $w_n(\#A)$ of words of length $n$ with respective number of $A$'s is thus:
$$\begin{align}
w_n(\#A) 
&:= \frac{(\#A+\#B+\#C+\#D)!}{\#A!\cdot\#B!\cdot\#C!\cdot\#D!} \\
&= \frac{n!}{\#A!^2(n/2-\#A)!^2}
\end{align}$$
Summing over all possible $a=\#A$:
$$\begin{align}
W_n &:= \sum_{a=0}^{n/2} w_n(a) \\
&= \sum_{a=0}^{n/2} \frac{n!}{a!^2(n/2-a)!^2}
\end{align}$$
$$\begin{align}
W_{2n} &= \sum_{k=0}^n \frac{(2n)!}{k!^2(n-k)!^2} \\
&= \sum_{k=0}^n \frac{(2n)!}{n!^2} \binom{n}{k}^{\!2}\\
&= \binom{2n}{n}\sum_{k=0}^n \binom{n}{k}^{\!2} 
= \binom{2n}{n}^{\!2}\\
\end{align}$$
For the last step and
$$\sum_{k=0}^n \binom{n}{k}^{\!2} = \binom{2n}{n}$$
see this question for example.
