Number of permutations of a set which contains 55 elements I am to find the number of permutations of a set which contains $55$ elements and the permutations meet following requirements:
$$\forall_{i\in \{1, 2, ..., 55\}} f(i)\neq i \wedge f \circ f = id$$
 A: Any permutation is a (unique) product of cycles that commute with each other. Since you want $f^2=id$, the only cycles allowed are transpositions. Since you want $f(i)\ne i$, all elements should appear in one transposition. 
This requires taking the elements in pairs. As the number of elements, 55, is odd, there will always be an element missing; any $f$ with $f^2=\text{id}$ has a fixed point. Thus, no such $f$ exists. 
When the number of elements is even, say $2n$ the pairing is obviously possible. The number of such pairings is given by
$$
\frac{(2n)!}{2^n\,n!}.
$$
For $f^3=\text{id}$, one would have to work with triples instead of pairs: the only cycles that are allowable are triples. So the problem becomes in how many way you can divide your elements in groups of $3$. Of course if you still what $f$ with no fixed elements (which is good, because otherwise the combinatorics become more complicated), then you need the number to be a multiple of $3$, say $3n$, and the number of possible $f$ will be 
$$
\frac{2^n(3n)!}{3^n\,n!}.
$$
The "new" $2^n$ in the numerator accounts for the fact that for every group of 3 elements there are two possible 3-cycles. 
