= Solution
{c}
For $0\le k\le\lfloor n/2\rfloor$, first select an ordered list of $2k$ distinct entries. There are $n!/(n-2k)!$ lists. Pair consecutive entries to obtain
$$
(a_1a_2)(a_3a_4)\cdots(a_{2k-1}a_{2k}).
$$
Each product of $k$ disjoint <transpositions> is counted $2^k k!$ times: each pair has two orientations, and the $k$ pairs can be ordered in $k!$ ways. There is no other duplication, by uniqueness of the <disjoint cycle decomposition>. Therefore
$$
\boxed{|T_k|=\frac{n!}{2^k k!(n-2k)!}.}
$$
If $2k>n$, there are no such <permutations>. For $k=0$, $T_0=\{e\}$.
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