Suppose we have $n$ elements, assume there is a permutations over $k$ elements among the $n$ elements so $n-k$ are fixed. Let that the permutation over the k elements is represented by permutation cycles so the length of all permutation cycles $=k$.
As an example: Suppose we have the following permutation
$$ x = \left( {\begin{array}{c}
x_1 = \left( {\begin{array}{c}
1 \\
2 \\
\end{array} } \right) \\
x_2 = \left( {\begin{array}{c}
3 \\
4 \\
5 \\
\end{array} } \right) \\
x_3 = \left( {\begin{array}{c}
6 \\
7 \\
\end{array} } \right) \\
8 \\
9 \\
\vdots \\
15 \\
\end{array} } \right)$$
My question: What is the number of permutations we can construct from the $n$ elements where each permutation should consists of the same cycles type?
Addition: I know that the number of $k-$cycles in the symmetric group $S_n$ is $\binom{n}{k}(k-1)!$ but I don't know what to do for the constraint asking that each permutation cycle has the same length in all permutations!
Best Answer
Hint: The number of distinct $k$-cycles is $P^n_k\cdot \dfrac 1k=\dfrac{n!}{(n-k)!}\cdot \dfrac 1k$.
To do your example, we would get, for permutations of type $(2,3,2)$ in $S_{15}$:
$P^{15}_2\cdot \dfrac 12\cdot P^{13}_3\cdot \dfrac 13\cdot P^{10}_2\cdot \dfrac 12=105\cdot572\cdot45=2702700$.
Now I need to divide by $2$, since I have double counted the two $2$-cycles.
So $\dfrac12\cdot2702700=1351350$.
See here, or here, for a good explanation.