Theorem 1.1

For integers $k$ and $n$ such that $0\le k\le n$, the number of $k$-permutations of a set of $n$ objects is $P(n,k)=\frac{n!}{(n-k)!}$^Definition

Proof

First suppose that $k=0$.
Then $P(n,0)$ and $\frac{n!}{(n-0)!}=\frac{n!}{n!}=1$, and hence the result holds
Now assume that $1\le k$, and consider the following counting procedure:

1. Choose an element for the first position ($n$ choices)
2. Choose an element for the second position ($n-1$ choices)
   $\dots$
3. Choose an element for the $k^{th}$ position ($n-k+1$ choices)
   So, by the Multiplication Principle, $P(n,k)=n(n-1)\dots (n-k+1)$
   $\frac{n!}{(n-k)!}=\frac{n(n-1)\dots (n-k+1)(n-k)!}{(n-k)!}=n(n-1)\dots (n-k+1)$, and hence the result holds true

Note that $P(n,n)=n!$