# 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:

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

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:

${}_{5}P{}_{3}=5\times 4\times 3=60$

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:

${}_{5}P{}_{3}=5\times 4\times 3=60$

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

${}_{8}P{}_{4}=8\times 7\times 6\times 5=1680$

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