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.
Best Answer
Algebraic proof. Your identity is equivalent to (note that the upper limit should be $k$ and the lower limit is $0$ because we moved $\binom{n}{2k}2^{2k}$ to the left hand side): $$\sum_{q=0}^k \binom{n}{q}\binom{n-q}{2k-2q}2^{2(k-q)} = \binom{2n}{2k}.$$ Now we have that $$\begin{align*} \sum_{q=0}^k \binom{n}{q}\binom{n-q}{2k-2q}2^{2(k-q)}&= \sum_{q=0}^k \binom{n}{q}2^{2(k-q)}[z^{2k-2q}] (1+z)^{n-q} \\ &=[z^{2k}]4^k(1+z)^n\sum_{q\geq 0} \binom{n}{q} \frac{z^{2q}}{4^q(1+z)^{q}} \\ &=[z^{2k}]4^k(1+z)^n\left(1+\frac{z^2}{4(1+z)}\right)^n\\ &=[z^{2k}]4^k\left(1+z+\frac{z^2}{4}\right)^n\\ &=[z^{2k}]4^k\left(1+\frac{z}{2}\right)^{2n}=\binom{2n}{2k}. \end{align*}$$