Multiple correct question based on permutation group 
I try to solve this the number of permutation in $S_{n}$ for $n\geq 4$ which are product of two disjoint $2$-cycles is $\frac{n(n-1)(n-2)(n-3)}{8}$.
So after putting $n=5$ and $n=4$ I get different answer. Am I right ?
Please help how to solve this. Thanks.
 A: There is an ambiguity in the question. If an element remains fixed, does that count as a 1-cycle?
If we assume it does, then for $n=4$ the only possibilities are type $(12)(34)$ (3 possibilities) or type $(123)(4)$ (8 possibilities, 4 ways of choosing the 1-cycle, then 2 ways of choosing the 3-cycle). So we get $a_4=11$.
For $n=5$ the only possibilities are type $(123)(45)$ (10 ways of choosing the elements in the 3-cycle, then 2 possibilities for each choice, total 20), or type $(1234)(5)$ (5 ways of choosing the 1-cycle, then 6 ways of ordering the elements in the 4-cycle). So $a_5=50$.
If we don't count 1-cycles, then:
For $n=4$ the only possibilities are $(12)(34),(13)(24),(14)(23)$. So $a_4=3$.
For $n=5$ the only possibilities are type $(123)(45)$ and type $(12)(34)$. The first has 20 possibilities (10 ways of choosing the elements in the 3-cycle and 2 possibilities for each choice). The second has 15 possibilities (5 ways of choosing which element remains fixed, then 3 $a_4$ possibilities). Hence $a_5=35$.
Presumably we are meant to adopt the first interpretation (count 1-cycles).
