# Permutations

When the elements of a set are arranged in a definite order, the arrangement is called a permutation of the elements.  The number of permutations of $n$ objects is $n!$

The number of possible orderings of $m$ objects taken from a set of $n$ is given by:

$\begin{array}{l}{}_{n}P{}_{m}=n×\left(n-1\right)×\left(n-2\right)×\cdots ×\left(n-m+1\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{n!}{\left(n-m\right)!}\end{array}$

That is, count backwards starting from $n$ , writing down the numbers as you count, until you've written down $m$ numbers. Then multiply them all together.

Example:

Suppose you're a television programmer, and you have five half-hour shows to choose from, but only three time slots. How many different programs are possible?

Using the permutations formula, we have:

${}_{5}P{}_{3}=5×4×3=60$

To see why this works, name the shows A, B, C, D, and E, and make a list:

 ABC ABD ABE ACB ACD ACE ADB ADC ADE AEB AEC AED BAC BAD BAE BCA BCD BCE BDA BDC BDE BAC BAD BAE CAB CAD CAE CBA CBD CBE CDA CDB CDE CEA CEB CED DAB DAC DAE DBA DBC DBE DCA DCB DCE DEA DEB DEC EAB EAC EAD EBA EBC EBD ECA ECB ECD EDA EDB EDC

In this case, there are $5$ choices for the first program, $4$ choices for the second program, and $3$ choices for the last program. So the answer is:

${}_{5}P{}_{3}=5×4×3=60$

If there were $8$ programs and $4$ time slots, we would have:

${}_{8}P{}_{4}=8×7×6×5=1680$