Permutation $ \pi$ has a signature $2^43^5$ (it contains $4$ cycles of length $2$ and $5$ cycles of length $3$).
Find number of permutations $ \delta $ such that $\delta ^4 = \pi$
My observation Cycle of length $3$: $$\pi = [a,b,c]$$ $$\pi^2 = [a,c,b]$$ $$\pi^3 = [a][b][c]$$ $$\pi^4 = [a,b,c]$$
Moreover for cycle of length $2$: $$\pi = [a,b]$$ $$\forall_k \, \pi^2 = \dots = \pi^k = [a,b]$$ So it means that number of these permutations is just all possible permutations with given signature? So it is $$\binom{23}{2}\binom{21}{2}\binom{19}{2}\binom{17}{2}\binom{15}{3}\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3} \cdot (3-1)!^5 \cdot (2-1)!^4 $$