## Thinking Mathematically (6th Edition)

Permutation problems are arrangement problems which can be solved by observing a successive list of questions: 1. In how many ways can we choose the 1st object? Answer: $n$. 2. In how many ways can we choose the 2nd object? Answer: $n-1$. ... r. In how many ways can we choose the rth object? Answer: $n-r+1$. The total number of ways is, using FCP, $n(n-1)n-2)\cdot...\cdot(n-r+1)$, The formula for ${}_{n}P_{r}$, (the number of ordered sequences taking r objects from n, each not being used more than once.), leads to the same expression : ${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!}=$ $=\displaystyle \frac{n(n-1)\cdot...\cdot ...(n-r+1)(n-r)!}{(n-r)!}$ ... after reducing $(n-r)!$, ... $= n(n-1)(n-2)...(n-r+1)$ The statement is true.