A permutation $\sigma$ of $\{1,2,\dots,14\}$ has $2$ cycles of length $4$ and $2$ cycles of length $3$. How many permutations $\pi$ are there st $\pi^2=\sigma$?
So $\pi$ could be made of $1$ cycle of length $8$ and $2$ cycles of length $3$ or $1$ cycle of length $8$ and $1$ cycle of length $6$. For the first option there are $4$ such permutation and for the second there are $4 \cdot 3 = 12$ permutations. But now we need to count how many ways we can distribute the $14$ elements for each of the two above options.
For the first case there are $\frac{14!}{8 \cdot 3 \cdot 3!}$ permutations and for the second one there are $\frac{14!}{48}$ permutations.
So in total we have $$ 4 \cdot \frac{14!}{8 \cdot 3 \cdot 3!} + 12 \cdot \frac{14!}{48}. $$ But this is on the order of $10^{10}$, which is very suspicious. What have I done wrong?
Let's first look at the $8$-cycle. An element of the first $4$-cycle can permute to any of the $4$ elements of the other $4$-cycle, and nowhere else. That "target" then must permute to the second element of the first $4$-cycle, which in turn must go to the second element of the second $4$-cycle. In short, there are $4$ possible $8$-cycles that work because your choice of the first "target" completely determines the $8$-cycle. (Thanks to @Daniel Schleper for the catch here.)
Similarly, if you have a $6$-cycle, there are $3$ possible $6$-cycles that work. There also is exactly $1$ pair of $3$-cycles that work. Therefore, there are $4 \cdot(3+1)=16$ permutations $\pi$ such that $\pi^2 = \sigma$.