Factorials and Arrangements
This page follows the JoeCMath video Factorials & Arrangements in 4 Min. The video introduces factorial notation, shows how quickly factorials grow, and connects the notation to counting arrangements.
Main Idea
For a positive integer $n$,
\[n! = n(n-1)(n-2)\cdots 3\cdot 2\cdot 1.\]$n!$ is read as “$n$ factorial.” It is especially useful when you are arranging distinct objects and order matters.
Quick Factorial Calculation
The video calculates $5!$ by multiplying from $5$ down to $1$:
\[5! = 5\cdot 4\cdot 3\cdot 2\cdot 1.\]So
\[5! = 120.\]Why $0!=1$
The video also points out the special definition
\[0! = 1.\]This may feel strange at first because there are no positive factors to multiply. In arrangement language, $0!$ counts the one way to arrange nothing: the empty arrangement. That definition also keeps factorial formulas working cleanly when a count reaches zero.
Factorials Count Arrangements
The arrangement idea comes from filling positions one at a time. If you have $n$ distinct objects, then:
- the first position has $n$ choices,
- the second position has $n-1$ choices,
- the third position has $n-2$ choices,
- and the pattern continues until the last position has $1$ choice.
Multiplying those choices gives
\[n(n-1)(n-2)\cdots 3\cdot 2\cdot 1 = n!.\]That is why $n!$ counts the number of ways to arrange $n$ distinct objects.
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 |
|---|---|
| $ABC$ | $ACB$ |
| $BAC$ | $BCA$ |
| $CAB$ | $CBA$ |
The table shows the same count as the factorial: there are $6$ possible arrangements.
Extending The Pattern
The video then extends the same reasoning to an arbitrary set of $n$ distinct objects. The first object has all of the choices available, and each position after that has one fewer choice left.
For $4$ distinct objects, the number of arrangements is
\[4! = 4\cdot 3\cdot 2\cdot 1 = 24.\]For $n$ distinct objects, the number of arrangements is
\[n!.\]Timestamp Guide
More JoeCMath Combinatorics
The video description points to the JoeCMath combinatorics playlist: watch the playlist on YouTube.