A permutation rearranges a set. For example, you could rearrange $($A$,$ B$,$ C$,$ D$)$ to be $($B$,$ A$,$ C$,$ D$)$ by switching the first two items. Alternatively, you could imagine an equilateral triangle with the corners labeled $1,2,$ and $3,$ and you could rotate it so that $1->2,2->3,$ and $3->1.$

In general, if S is a set, and $\backslash$pi is a permutation of S$,$ then for each element s in S$,\backslash$pi bracket$($s$)=$t for some t in S$.$ Because this relates to permutations, you do not want two different elements mapped to the same place, which would correspond to folding the triangle in half, rather than rotating it$.$ Thus, you require that if a$,$b in S$,$and$$a$!=$b$,$ then $\backslash$pi bracket$($a$)!=\backslash$pi bracket$($b$).$

Suppose that S is finite, and that $\backslash$pi bracket$($s$)!=$s$,$ but $\backslash$pi bracket$(\backslash$pi bracket$($s$))=$s for all s in S$.$ What can be said about the number of elements in S$?$