fixed length of permutaions cycles

Question

How much permutations has only 10 cycles, but three of them has length 3 and seven of them has length 7?

Answer 1

The answer is $$\frac{(2!)^3(6!)^7}{3!7!}\binom{58}{3,3,3,7,7,7,7,7,7,7}=\frac{58!}{3^3 7^7(3!)(7!)}$$ which is an enormous number. The set of 58 elements is broken into 10 disjoint subsets of the indicated sizes, this number is divided by $3!7!$ to account for the fact that the order of the cycles doesn't matter, after which we have to take into account the ordering of the elements within the cycle. There are $2!$ distinct orderings for length 3 and $6!$ for length 7.