Permutations Intro
This page follows the JoeCMath video Permutations Intro in 4 Min: 2 Easy to Follow Examples. The video introduces permutations as ordered arrangements, then uses two examples to show why the order of the choices matters.
Main Idea
A permutation counts ordered arrangements. The order matters because changing who goes in which position creates a different outcome.
In the video, this shows up in examples like:
- who gets $1\text{st}$, $2\text{nd}$, and $3\text{rd}$ place in a pizza eating competition,
- who becomes president and who becomes vice president.
Those are ordered roles. Switching the people around changes the result.
Watch this section: loose permutation definition at 0:00.
The Permutation Formula
The video uses the notation $nPr$, which is often written in LaTeX as ${}_nP_r$. It means: arrange $r$ objects from a set of $n$ objects.
The formula is
\[{}_nP_r = \frac{n!}{(n-r)!}.\]The same idea can also be read as a decreasing product:
\[{}_nP_r = n(n-1)(n-2)\cdots (n-r+1).\]The product stops after $r$ factors because only $r$ positions are being filled.
Watch this section: general permutation formula at 1:02.
Pizza Competition Example
The first example in the video counts possible $1\text{st}$, $2\text{nd}$, and $3\text{rd}$ place outcomes for $5$ competitors.
For the direct counting view:
- there are $5$ choices for $1\text{st}$ place,
- then $4$ choices for $2\text{nd}$ place,
- then $3$ choices for $3\text{rd}$ place.
So the number of ordered outcomes is
\[5\cdot 4\cdot 3 = 60.\]Using permutation notation, this is
\[{}_{5}P_{3} = \frac{5!}{(5-3)!}.\]Then
\[{}_{5}P_{3} = \frac{5!}{2!} = \frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1} = 5\cdot 4\cdot 3 = 60.\]The formula gives the same result as the direct choice count.
Watch this section: pizza eating competition example at 0:05.
President And Vice President Example
The second example counts ways to choose a president and a vice president from $8$ students.
There are two different positions, so order matters. If Alice is president and Jordan is vice president, that is different from Jordan as president and Alice as vice president.
The direct counting view is:
\[8\cdot 7 = 56.\]There are $8$ choices for president. After one student is chosen, there are $7$ choices left for vice president.
Using the permutation formula,
\[{}_{8}P_{2}=\frac{8!}{(8-2)!}.\]Then
\[{}_{8}P_{2}=\frac{8!}{6!} = \frac{8\cdot 7\cdot 6!}{6!} = 8\cdot 7 = 56.\]Again, the formula matches the direct reasoning.
Watch this section: president and vice president example at 2:07.
Why ${}_nP_n=n!$
Near the end, the video points out that arranging all $n$ objects is the same as a factorial.
If $r=n$, then
\[{}_nP_n = \frac{n!}{(n-n)!}.\]Since $n-n=0$ and $0!=1$,
\[{}_nP_n = \frac{n!}{0!}=\frac{n!}{1}=n!.\]So
\[{}_nP_n=n!.\]That connects this video back to factorials: factorials count arrangements when every object is used.
Watch this section: why $nPn=n!$ at 3:27.
Timestamp Guide
Related Combinatorics Work
For the factorial background behind the formula, use the companion guide for factorials and arrangements.
The video description also points to the JoeCMath combinatorics playlist: watch the playlist on YouTube.