TITLE: (combinatorial and algebraic) proof that $\sum_{q=1}^n \binom{n}{q}\binom{n-q}{2k-2q}2^{2(k-q)} = \binom{2n}{2k} - \binom{n}{2k}2^{2k}$
QUESTION [9 upvotes]: I arrived at the following identity throught an attempt at a homework problem: the LHS is my solution; the RHS, the book's one. And as far I have tested, they are equivalent.
$$\sum_{q=1}^n \binom{n}{q}\binom{n-q}{2k-2q}2^{2(k-q)} = \binom{2n}{2k} - \binom{n}{2k}2^{2k}$$
The combinatorial meaning of this identity is rather intuitive as per the homework's verbatim:
"Find the number of ways of forming a group of $2k$ people from $n$ couples where $k,n \in \Bbb N$ and $2k \leq n$ so that at least one couple is included in such group".
That being said, the combinatorial meaning of the LHS (and even the RHS) is rather clear: it iterates over $q = \{1,2,...,n\}$ couples, selects $2k-2q$ couples from the remaining $n-q$ and then picks one person from each of these $2k-2q$ couples,  which can be done in $2^{2k-2q}$ ways, to form the group. The RHS does this by composition: removing those groups that do not contain one couple.
However, while I know what the LHS is supposed to do, I don't know how it works. I know that the groups that do not contain 1 couple is given by $\binom{n}{2k}2^{2k}$ (after all, this is the case where $q=0$), but I have little idea as to how the $\binom{2n}{2k}$ term counts the "rest". And I have way fewer ideas about how would I prove this thing algebraically.
Later edit: I realized that the $\binom{2n}{2k}$ term simply counts the number of ways of grouping $2k$ people from the overall $2n$ people of the $n$ couples. So ofc all you'd need to do after that is remove those groups that do not contain 1 couple.

REPLY [4 votes]: Here's a "snake oil" proof of the generalization noted by @darijgrinberg:
\begin{align}
\sum_{k \ge 0} \sum_{q=0}^n \binom{n}{q}\binom{n-q}{k-2q}2^{k-2q} z^k
&= \sum_{q=0}^n \binom{n}{q} \sum_{k=2q}^{n+q} \binom{n-q}{k-2q}2^{k-2q} z^k \\
&= \sum_{q=0}^n \binom{n}{q} \sum_{k=0}^{n-q} \binom{n-q}{k}2^k z^{k+2q} \\
&= \sum_{q=0}^n \binom{n}{q} z^{2q} \sum_{k=0}^{n-q} \binom{n-q}{k}(2z)^k \\
&= \sum_{q=0}^n \binom{n}{q} z^{2q} (1+2z)^{n-q} \\
&= (1+2z)^n \sum_{q=0}^n \binom{n}{q} \left(\frac{z^2}{1+2z}\right)^q \\
&= (1+2z)^n \left(1+\frac{z^2}{1+2z}\right)^n \\
&= (1+2z)^n \left(\frac{(1+z)^2}{1+2z}\right)^n \\
&= (1+z)^{2n} \\
&= \sum_{k \ge 0}\binom{2n}{k} z^k
\end{align}
Hence
$$\sum_{q=0}^n \binom{n}{q}\binom{n-q}{k-2q}2^{k-2q} = \binom{2n}{k}.$$