Solution
= Solution
The <product of two transpositions> has three possible forms. Equal transpositions produce the identity, transpositions meeting in one letter produce a <three-cycle>, and disjoint transpositions produce two disjoint two-cycles. There are two three-cycles on each chosen triple and three pairings of each chosen four-set. Therefore the number of distinct permutations representable this way is
$$
\boxed{1+2\binom n3+3\binom n4}.
$$