Let $a_n$ denote the number of those permutations $\sigma$ on $\{1,2,3....,n\}$ such that $\sigma$ is a product of exactly two disjoint cycles. Then    $a_5 = 50$ $a_4 = 14$ $a_5 = 40$ $a_4  = 11$   I tried specifically for $a_5$  and $a_4$ with a little bit of calculations. But I want to know about a formula for $a_n$ with less calculations
Let $a_n$ be the number of those permutation $\sigma $ on $\{1,2,...,n\}$  such that $\sigma $ is a product of exactly two disjoint cycles. Then find $a_4$ and $a_5$.   Calculating $a_4$: Possible cases which can happen are $(12)(34),(13)(24),(14)(23)$, any cycle of the form $(123)$ or $(12)$ i.e. two-cycles and three cycles thus we have in total $3+\frac{1}{3}4P_3+\frac{1}{2}4P_2=3+8+6=17$ but the correct answer is given to be either $11$ or $14$.  Where am I wrong? Please help.
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