Permutations of distinct objects

Viewed as a function, a permutation of a finite set $S=\{x_1,x_2,\ldots,x_n\}$ is a bijective mapping of $S$ onto itself; the number of permutations of $S$ therefore equals the total number of bijections from $S$ to $S$, which is $n!$.

Theorem 4.1. If $n,r$ are positive integers with $r\le n$, the number of permutations of $n$ distinct objects taken $r$ at a time is

Proof idea. A permutation of $r$ objects out of $n$ fills $r$ positions in a row using $n$ distinct objects: the first position has $n$ choices, the second has $n-1$ remaining choices, the third has $n-2$, ..., and the $r$th position has $n-(r-1)$ choices left. By the rule of product, $^nP_r=n(n-1)(n-2)\cdots(n-r+1)$.

Theorem 4.2. For $n\ge1$, $0\le r\le n$,

Proof. Multiply and divide Theorem 4.1's product by $(n-r)!$:

Boundary values. For a positive integer $n$ and non-negative integer $r$:

So arranging all $n$ distinct objects in a row is $^nP_n=n!$ ways, matching the earlier direct count. …