# Introduction to permutations (a mathy word)

Updated: Jan 5

By Sumay McPhail

In this lesson we are going to learn about permutations and combinations. These are helpful in solving the number of ways that a particular list can be created such as how many combinations there are to a lock or outfits we can make out of a set number of clothes. I hope you enjoy!

A permutation is a list in which the order matters such as a combination. We would not be able to open a lock if we put the code in backwards, which is why the number of possible combinations in a lock is called a permutation. For example, suppose that we have a locker code that is three numbers long. Suppose further that it only includes possible numbers 1, 2, and 3 … and we can only use each number once. How many permutations are there? Possible answers to this include 123, 132, 213, 231, 312, 321. That is six permutations, and not hard to figure out by writing them down. But what if the lock allows all the numbers 1 through 10? For this we need a formula which is called a factorial “!”. In this case the factorial of 3 which is written as “3!” and means 3 x 2 x 1 = six. Same answer!

Our next permutation is going to allow replacement; which is how most locks work. Suppose we have the same lock as before but this time the lock allows you to use the same number more than once. We could write this out as 111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333. Now that’s a lot more combinations! No way I want to do that with more digits like 10...so we need a way to solve permutations with replacement formulas. This formula uses the “power” sign “^” so which in this case looks like 3^3. That means 3 to the power of 3 or 3 x 3 x 3. Notice that this delivers 27 combinations compared to just 6 from 3! which delivers just 3 x 2 x 1 combinations. I like that lock better. Much more secure!

We can use permutations to solve real world problems like how many outfits we can make given different sets of clothes. For example, suppose we want to know how many combinations we can make from two shirts, three pants, and two coats. Let’s give the clothes letter names: Our shirts can be named A and B. Our pants can be named X, Y, and Z. And our coats can be named C and D. We can write out these combinations as AXC, AXD, AYC, AYD, AZC, AZD, BXC, BXD, BYC, BYD, BZC, BZD. That’s a total of 12 combinations. But what formula can we use to solve this in case we buy more clothes (I have way more clothes). Let's assume we don’t care about fashion so we can wear any combination of shirt, pants and coat (just focus on the math not the image). That means we are looking at a permutation with replacement. For this problem we know we want to use the power formula 2 x 3 x 2 which gives us the same answer of 12!

In this lesson we learned about permutations which are helpful in trying to find the number of combinations that we can make from a given set of numbers, letters, clothes, or anything else. We provided formulas because trying to write out combinations can be very tedious when the numbers get large. Try out the questions below to see if you got it!

**Question 1: What does 10! Mean?**

A. 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800

B. 10! = 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,000,0000,000

C. 10! = A very excited ten!

D. 10! = 10 x 10 = 100

**Question 2: What does 10^10 Mean?**

A. 10^10 = 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,000,0000,000

B. 10^10 = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800

C. 10^10 = A very excited ten!

D. 10^10 = 10 + 10 = 20

**Question 3. ****How many combinations of clothes can you make from 10 shirts, 5 pants, and 3 coats?**

A. 10 x 5 x 3 = 150

B. 10 + 5 + 3 = 18

C. 10! x 5! x 3! = a lot!

D. Depends on what colors they are.