Textbook

Permutations and combinations questions ask you about the number of possible ways an outcome could occur. We’ll break down these complicated statistics concepts and teach you a few tricks to solve these questions.

The initial question you should ask yourself for any permutation or combination question should be: *“How many decisions do I have to make and how many options are there per decision?”*.

Let’s say you are at a restaurant and have to choose food and a drink. There are five drinks and 10 food items on the menu. How many total combinations are there with exactly one drink and one food item?

To solve this, we can write out two blanks (one for each decision) and fill in the slots with the number of options per decision (5 and 10). Then we’ll multiply the numbers in the blanks to get the final answer of $50$.

$5 ×10 =50$

This same technique works for all permutation questions, but sometimes it takes a little thought to figure out how to fill in the blanks.

Five kindergarteners are lining up for recess. How many ways could they line up?

How many decisions must be made, and how many options per decision? Well, there are five kids, so there are five decisions. For the first decision (the first kid in line) we could choose any of the five kids. For the second decision (the second kid in line), we have one less kid to choose from (since one of the kids is already first in line), so the number of options for this decision is only four. This pattern continues all the way down to one kid to choose for the final slot (since with five kids and five spots, every kid will get a spot). We multiply each slot to get a final answer of $120$.

$5 ×4 ×3 ×2 ×1 =120$

Great! Now let’s try a more complicated permutation question.

Greg needs to get five tasks done this Saturday before lunch.

- Go grocery shopping
- Mow the lawn
- Clean the cat’s litter box
- Visit grandma
- Do the laundry
Greg wants to mow the lawn first, and wants to go grocery shopping before visiting grandma so we can bring her some vegetables. How many different ways can Greg plan his tasks for the day?

Try it yourself, and then check the answer and explanation below.

(spoiler)

There are 12 unique permutations that satisfy the question constraints.

The first thing to consider is that Greg must mow the lawn first. The question clearly states this, and although it’s included in the task list, it’s essentially a fixed outcome as it will always be first, meaning that it isn’t actually part of the permutation. This leaves Greg with four tasks that can be arranged in *almost* any way, keeping in mind that Greg must go shopping before visiting grandma.

Let’s start by calculating how many permutations we have total, ignoring the shopping-before-grandma rule. There are four decisions to make and four options, so our equation looks like this:

$4 ×3 ×2 ×1 =24$

There are 24 ways to arrange these tasks if Greg doesn’t worry about giving grandma vegetables. Out of these 24 scenarios, what percentage of them don’t satisfy that constraint? Well, in half the cases he would be going shopping before visiting grandma, and in half the cases he would be going shopping after visiting her. Therefore, 50% of the permutations don’t satisfy the constraint, and that leaves us with the other 50% of the 24 permutations that are valid, i.e. 12 ways that Greg can plan his morning.

Order matters for permutations (e.g. the first kid in line followed by a different kid second) whereas it does not matter for combinations. Here’s an example combination problem:

How many pairs of news anchors could be picked from six news anchor applicants?

It doesn’t matter who is picked first or second. We simply want to know how many pairs exist. You should first start this problem just as you start a permutation problem. How many decisions need to be made, and how many options are there? There are two decisions (since each pair includes two people), with six options for the first person and five options for the second person. If this were a permutation problem (e.g. how many ways to order a 2-person line from 6 people total), we’d be finished. However, the order doesn’t matter since we’re asked about the total combinations. Picking Alex 1st and Blair 2nd is the same combination as picking Blair 1st and Alex 2nd.

One final step is needed to eliminate the duplicate combinations. We have two slots, and that means we need to divide by $2!$. That exclamation mark means factorial, and $2!$ is equal to $2∗1=2$. So, dividing by $2$ will eliminate the duplicates.

$2 ×1 6 ×5 =15$

The final answer is that $15$ pairs of news anchors could be made from a group of six people.

Let’s do another combinations question.

How many groups of four coins can be pulled from a coin purse that contains nine unique coins from all around the world?

As always, try it yourself first, and then check your work!

(spoiler)

There are 126 groups of 4 coins that can be selected.

Let’s walk through it. We have four decisions to make, which means four slots in our equation. For the first slot, there are nine coins to choose from, and for each slot after that, we’ll have one less option.

$9 ×8 ×7 ×6 =3024$

The order we choose the four coins does not matter, since the question asks how many “groups” of four coins can be made. Selecting a [penny, nickle, dime, quarter] group shouldn’t be double-counted if you select a [quarter, dime, nickel, and penny] group. This is a combination question and not a permutation question. We’ve overcounted and need to eliminate the duplicate groups!

We need to divide by “the-number-of-slots factorial”. There are four slots, so we divide by $4!$.

$4 ×3 ×2 ×1 9 ×8 ×7 ×6 =126$

Wrapping it up, 126 (unique) groups of 4 coins can be selected.

The methods we’ve described above are effectively the practical application of mathematical equations for permutations and combinations. If you prefer to solve questions mathematically, this section is for you.

Remember, use *permutations* when the order of elements is important, like finding the number of ways people could queue up in a line.

Variables:

- $n$ is the total number of options available
- $k$ is the number of times you have to choose

$Permutations=(n−k)!n! $

Let’s solve the following question using this equation.

Five kindergarteners are lining up for recess. How many ways could they line up?

Which variables match $n$ and $k$? Use those to create and solve the permutations equation.

(spoiler)

$n=5$

$k=5$

$Permutations=(n−k)!n! =(5−5)!5! =120$

There are 5 kindergarteners, which means we have 5 total possible options to choose from, giving us $n=5$. We need to have all 5 kids in line, which means that $k=5$ as well.

Here’s the full math equation:

$Permutations=(n−k)!n! =(5−5)!5! =0!5! =15∗4∗3∗2∗1 =120$

Using the equation will always result in the same answer as the permutations “blanks” method we discussed earlier. Feel free to solve any problem either way! Just make sure to first identify if this is a permutations or combinations question.

Remember, use *combinations* when the order of elements is *not* important, like finding the number of ways you could pick marbles from a jar containing multiple colors.

Variables:

- $n$ is the total number of options available
- $k$ is the number of times you have to choose

$Combinations=k!(n−k)!n! $

Let’s solve the following question using this equation.

How many pairs of news anchors could be picked from six news anchor applicants?

Which variables match $n$ and $k$? Use those to create and solve the combinations equation.

(spoiler)

$n=6$

$k=2$

$Combinations=k!(n−k)!n! =2!(6−2)!6! =15$

There are 6 total news anchors to choose from, so $n=6$. We want to find the pairs of people, i.e. groups of 2 people, so $k=2$.

Here’s the full math equation:

$Combinations=k!(n−k)!n! =2!(6−2)!6! =2!∗4!6! =(2∗1)∗(4∗3∗2∗1)6∗5∗4∗3∗2∗1 =2∗16∗5 =15$

Did you notice how the $4∗3∗2∗1$ in the denominator of the fraction canceled out the same $4∗3∗2∗1$ from the top? This frequently occurs in combination and permutation equations and can help you simplify your math, so check to see if there are shared terms you can cancel before doing the multiplication.

Using the equation will always result in the same answer as the combinations “blanks” method we discussed earlier. Feel free to solve any problem either way! Just make sure to first identify if this is a permutations or combinations question.

Although it is tested infrequently, you might come across a strange-looking notation that looks like a fraction without a dividing line. This is called **n Choose k** notation, or simply **nCk**. Formally, it is known as the binomial coefficient.

$(kn )$

This describes the total number of results of a simple combination. The $n$ represents the total number of things to choose from, and the $k$ represents the number of things you’re choosing.

For instance, the notation $(35 )$ describes the total number of ways you could arrange 3 things from 5 total things, like the 10 possible arrangements of 3 paintings from a selection of 5 total paintings. It could also be written as 5C3, or 5 choose 3.

$(35 )=3!(5−3)!5! =(3∗2∗1)(2∗1)5∗4∗3∗2∗1 =5∗2=10$

Let’s think about the coin question in the previous section.

How would you write the equation using nCk notation?

$4×3×2×19×8×7×6 =126$

A. $(19 )/(44 )$

B. $(49 )$

C. $(19 )/(14 )$

D. $(99 )$

Do you know the answer?

(spoiler)

Answer: B. $(49 )$

The trick to solving this question quickly is realizing:

- The numerator starts with $9$, so $n$ is likely $9$
- The denominator doesn’t have $5$, so $k$ is likely $5$

If you do the math, you will see it checks out!

Typically, a question will appear to be a regular combination or permutation problem, and only the final answer choices will be written in the nCk format.

Here’s a video going through one of our practice questions to demonstrate these ideas in action:

Sign up for free to take 4 quiz questions on this topic

All rights reserved ©2016 - 2024 Achievable, Inc.