Permutation Notation

Finding Permutations

When working with permutations, the first thing to know is that they are different from combinations.

With combinations, order doesn't matter. Jaya, Zach, and Marlee is the same as Zach, Marlee, and Jaya. Either way, it's the same combination of people.

With permutations on the other hand, order matters. Jaya, Zach, and Marlee is a totally different permutation than Marlee, Jaya, and Zach.

Permutations show us how many possible ways there are do something.

Here's another example, based on a joke that math teachers like to tell:

Let's try a practice problem to see how we can calculate permutations.

To answer this question, we need to calculate how many permutations there are for the top 3 players given that we have 7 players overall. Remember, order matters.

Here's the general formula for calculating permutations: $$_nP_r = \frac{n!}{(n-r)!}$$

• $P$ stands for permutation.
• $n$ is the number of people or objects to choose from.
• $r$ is the number of spaces to fill.
• The $!$s in this equation don't mean that you need to read it really loudly. In math, exclamation points are the symbol for factorials.

To find a factorial, multiply the number before the exclamation point by every number less than it, all the way down to 1.

For example: $$\begin{align}5! &= 5\cdot 4\cdot 3\cdot 2\cdot 1 \\ &= 120\end{align}$$

Let's practice factorials before we bring our focus back to permutations.

Let's come back to our permutation problem now that we know what factorials are.

$$_7P_3 = \frac{7!}{(7-3)!}$$

(Video) The Counting Principle, Permutations, and Combinations

by Jack Walker

Jack Walker has a great video on the multiplication counting principle, permutations, and combinations.

A permutation tells you how many ways you can arrange $n$ things in $r$ slots.

The formula for finding how many permutations there are is: $$\boxed{\frac{n!}{(n-r)!}}$$ Remember, $n!=n\cdot (n-1)!\cdot (n-2)!\cdot (n-3)!\cdots 1$. If we look at this with an example, it will become clear the easiest way to calculate this. $$n=9,\;\;r=3\;\;\rightarrow\;\; \frac{9!}{(9-6)!}\;\;\rightarrow\;\; \frac{9\cdot8\cdot7\cdot\color{Red} {6\cdot5\cdot4\cdot3\cdot2\cdot1}} {\color{Red}{6\cdot5\cdot4\cdot3\cdot2\cdot1}}$$ All the numbers in red cancel out with each other. This means that the expression simply turns into: $$9\cdot8\cdot7 = 504$$ Instead of writing out that big fraction every time, we have a notation for this. Permutation Notation is: $$_nP_r$$ So with the example above it would be: $$_9P_3=\frac{9!}{(9-6)!}=9\cdot8\cdot7 = 504$$ For Permutation notation, $n$ is always the bigger number and $r$ is the smaller number. If $r$ is equal to $n$, then in the denominator you get $(n-n)!\;\;\rightarrow\;\;0!$. To make sense of this, $0! =1$. That means, if: $$n=r,\;\; _nP_n=n!$$ This makes sense because arranging $n$ things into $n$ slots is the same as arranging $n$ things in as many orders as you can, which is exactly what $n!$ does.

PERMUTATIONS

Permutations are used when you want to find how many different groups of things you can create out of a larger set of things, but the order matters. The easiest way to show what exactly you are looking for in a permutation is called Permutation Notation and looks like this: $$_n P_r$$ where $n$ is the number of objects you have in the entire set, and $r$ is the number of objects in each ordered group you are trying to find. In all cases, $n$ will always be the larger of the two numbers as it is the group that you are picking elements from. The equation for calculating the number of permutations is as follows: $$_n P_r=\frac{n!}{\left(n-r\right)!}$$ We’re just going to dive right into a problem in order to learn how to use this equation.

Let’s say that there is a local race track that has weekly auto races with $10$ people in each race. Each person that places in 5th place or higher earns points that go towards their standings for the entire season, with the higher placing racers getting more points. In any given race, how many different permutations of point-earning racers would there be?

First, you have to decide which number is $n$ and which is $r$. Here, $n = 10$ as that is the total number of racers, and $r = 5$, as that is the number of racers that can score points in each race. The equation would look like this with the numbers plugged in: $$_{10} P_5=\frac{10!}{(10-5)!}=\frac{10!}{5!}$$ To solve this, you can’t just just divide 10 by 5, then take the factorial. However, you also don’t need to find each factorial individually before you divide the numbers. You can cancel out factors that are in both the numerator and denominator. In this case, its $(5\cdot4\cdot3\cdot2\cdot1)$. $$\begin{align}\frac{10!}{5!}&=\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1}\\ &=10\cdot9\cdot8\cdot7\cdot6 \\ &=\boxed{30{,}240}.\end{align}$$

If you're still a little uncomfortable with the idea, here is a video that goes over both Permutations and Combinations in depth by Brandon Foltz on YouTube.