1.2.1 Permutations

Consider the selection of $k$ objects from a set of $n$ objects where order matters and repetition is not allowed.
Any such selection is called a $k$ -permutation. The total number of $k$ -permutations of a set of $n$ (distinct) objects is denoted by $P (n, k)$ .

For integers $k$ and $n$ such that $0 \leq k \leq n$ , the number of $k$ -permutations of a set of $n$ objects is $P (n, k) = \frac{n!}{(n - k)!}$ ###### 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 $1 \leq k$ , and consider the following counting procedure:

Choose an element for the first position ( $n$ choices)
Choose an element for the second position ( $n - 1$ choices)
$\dots$
Choose an element for the $k^{t h}$ 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 $P (n, n) = n!$

Example

How many 7 letter strings with no repeated letters can be formed from the English alphabet

The alphabet has 26 letters $∴ n = 26$
We want 7 letter strings $∴ k = 7$
$n - k = 19$
$P (n, k) = P (26, 7) = \frac{26!}{(26 - 7)!} = \frac{26!}{19!}$