 Solution

Write $n=2r$ and $k=2s$. Partition an $n$-element set into $r$ distinguished pairs, and let $\tau$ interchange the two elements of every pair. It induces an involution on the $k$-element subsets. Every nonfixed subset is paired with its distinct image, so nonfixed subsets contribute an even number to the binomial coefficient $\left(\genfrac{}{}{0ex}{}{2r}{2s}\right)$.

A fixed subset contains either both or neither of the elements of each distinguished pair. Since it contains $2s$ elements, it consists of exactly $s$ whole pairs. There are $\left(\genfrac{}{}{0ex}{}{r}{s}\right)$ such fixed subsets. Counting modulo $2$ therefore givesThe same involution explains part (i): a fixed subset cannot have odd size.