Factorials and Arrangements
Simple overview of factorials.
Main Idea for Factorials
For a positive integer $n$,
\[n! = n(n-1)(n-2)\cdots 3\cdot 2\cdot 1.\]$n!$ is read as “n factorial.”
A common first application is counting the ordered arrangements of a set of $n$ objects.
Simple Example (Writing out and finding value of $5!$)
To find the value of $5!$ lets write it out
\[5! = 5\cdot 4\cdot 3\cdot 2\cdot 1.\]Multiplying everything together gets us
\[5! = 120.\]Factorial Relationship with Arrangements
Suppose we have $n$ distinct objects and want to arrange them into $n$ positions:
- the first position has $n$ choices,
- the second position has $n-1$ choices (we remove one choice from the first position),
- the third position has $n-2$ choices (we remove two choices from the previous two positions),
- and the pattern continues until the $n^{\text{th}}$ position has $1$ choice (we remove $n-1$ choices from the previous $n-1$ positions).
If we then multiply all $n$ positions together we get
\[n(n-1)(n-2)\cdots 3\cdot 2\cdot 1 = n!\]So $n!$ is a quick way to find the number of ways we can arrange $n$ distinct objects in $n$ positions.
Arrangement Example With Three Objects
Suppose the objects are $A$, $B$, and $C$. There are $3$ choices for the first position, then $2$ choices for the second position, then $1$ choice for the last position:
\[3! = 3\cdot 2\cdot 1 = 6.\]| Arrangement | Arrangement |
|---|---|
| $ABC$ | $ACB$ |
| $BAC$ | $BCA$ |
| $CAB$ | $CBA$ |
Notice: The table that contains the unique arrangements of $A$, $B$, and $C$ contains $6$ arrangements, the same as $3!=6$.
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