Permutations and combinations | Arithmetic & algebra | Quantitative reasoning

Permutations and combinations questions ask you to count how many different ways an outcome can happen. In this lesson, you’ll learn a simple decision-based method (the “blanks” method) and then connect it to the standard formulas.

Permutations

For any permutation or combination question, start by asking:

“How many decisions do I have to make, and how many options are there for each 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?

You’re making two decisions:

Choose a drink (5 options)
Choose a food item (10 options)

Write one blank for each decision, fill in the number of options, and multiply:

$\underline{5} \times \underline{10} = 50$

This same technique works for all permutation questions. The main challenge is figuring out how many options belong in each blank.

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

You’re filling 5 spots in a line, so you have 5 decisions.

First spot: 5 choices
Second spot: 4 choices (one child is already used)
Third spot: 3 choices
Fourth spot: 2 choices
Fifth spot: 1 choice

Multiply the choices for each spot:

$\underline{5} \times \underline{4} \times \underline{3} \times \underline{2} \times \underline{1} = 120$

Now 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.

Start with the fixed condition: Greg must mow the lawn first. Since that task is always in the first position, it doesn’t create a choice. That leaves 4 tasks to arrange, with one additional rule: grocery shopping must happen before visiting grandma.

First, count the number of ways to arrange the remaining 4 tasks if you ignore the shopping-before-grandma rule:

$\underline{4} \times \underline{3} \times \underline{2} \times \underline{1} = 24$

Now apply the constraint. In these 24 arrangements, grocery shopping and visiting grandma can appear in either order:

In half the arrangements, shopping comes before grandma.
In the other half, grandma comes before shopping.

So only 50% of the 24 arrangements are valid:

Valid permutations: $\frac{1}{2} \times 24 = 12$

Combinations

Order matters for permutations (for example, who is first in line versus second). Order 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?

A “pair” means you’re choosing 2 people, and it doesn’t matter who is chosen first.

Start the same way you would for a permutation:

First choice: 6 options
Second choice: 5 options

That gives $6 \times 5$ ordered selections. But each pair is counted twice (Alex then Blair, and Blair then Alex). To remove the duplicates, divide by $2!$ .

That exclamation mark means factorial. Here, $2! = 2 \times 1 = 2$ .

$\frac{6 \times 5}{2 \times 1} = 15$

So, $15$ pairs of news anchors can 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.

You’re choosing 4 coins, so start by counting ordered selections:

First coin: 9 options
Second coin: 8 options
Third coin: 7 options
Fourth coin: 6 options

$\underline{9} \times \underline{8} \times \underline{7} \times \underline{6} = 3024$

But the question asks for “groups,” so order doesn’t matter. Selecting a [penny, nickle, dime, quarter] group shouldn’t be counted separately from selecting a [quarter, dime, nickel, and penny] group. Since order doesn’t matter, you’ve overcounted.

To remove duplicates, divide by “the-number-of-slots factorial.” There are 4 slots, so divide by $4!$ :

$\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

See how the number of slots in the divisor is the same as the number of slots in the numerator? They always match up for combination questions.

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

Equations for permutation and combination questions

The “blanks” method above is a practical way to apply the standard permutation and combination equations. If you prefer using formulas directly, use this section.

Permutations equation

Use permutations when the order of elements matters, like counting the number of ways people can line up.

Variables:

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

$Permutations = \frac{n !}{( n - k )!}$

Let’s solve the following question using this equation.

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

Which values match $n$ and $k$ ? Use them to create and solve the permutations equation.

(spoiler)

$n = 5$
$k = 5$

$Permutations = \frac{n !}{( n - k )!} = \frac{5 !}{( 5 - 5 )!} = 120$

There are 5 kindergarteners total, so $n = 5$ . You’re placing all 5 into the line, so $k = 5$ .

Here’s the full math equation:

$Permutations = \frac{n !}{( n - k )!} = \frac{5 !}{( 5 - 5 )!} = \frac{5 !}{0 !} = \frac{5 * 4 * 3 * 2 * 1}{1} = 120$

Sidenote

Zero factorial (0!) is equal to 1

It’s important to know that $0! = 1$ .

Think of it this way: there is only 1 way to group 0 things, meaning one group containing nothing.

Using the equation will always give the same answer as the permutations “blanks” method. Use whichever method you prefer, but always start by deciding whether the question is asking for permutations or combinations.

Combinations equation

Use combinations when the order of elements does not matter, like counting the number of ways to pick marbles from a jar.

Variables:

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

$Combinations = \frac{n !}{k ! ( n - k )!}$

Let’s solve the following question using this equation.

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

Which values match $n$ and $k$ ? Use them to create and solve the combinations equation.

(spoiler)

$n = 6$
$k = 2$

$Combinations = \frac{n !}{k ! ( n - k )!} = \frac{6 !}{2 ! ( 6 - 2 )!} = 15$

There are 6 total news anchors to choose from, so $n = 6$ . A pair is a group of 2, so $k = 2$ .

Here’s the full math equation:

$Combinations = \frac{n !}{k ! ( n - k )!} = \frac{6 !}{2 ! ( 6 - 2 )!} = \frac{6 !}{2 ! * 4 !} = \frac{6 * 5 * 4 * 3 * 2 * 1}{( 2 * 1 ) * ( 4 * 3 * 2 * 1 )} = \frac{6 * 5}{2 * 1} = 15$

Did you notice how the $4 * 3 * 2 * 1$ in the denominator canceled out the same $4 * 3 * 2 * 1$ from the top? This often happens in permutation and combination equations. Before you multiply everything out, look for shared terms you can cancel.

Using the equation will always give the same answer as the combinations “blanks” method. Use whichever method you prefer, but always start by deciding whether the question is asking for permutations or combinations.

n Choose k notation (nCk)

Although it’s tested infrequently, you might see a notation that looks like a fraction without a dividing line. It represents the combinations calculation, which is why it uses “C.”

$(k n) = (3 5) = \frac{5 * 4 * 3}{3 * 2 * 1}$

The notation $(3 5)$ means “5C3” or “5 choose 3.” It’s another way to write the number of ways to choose a group of $3$ from $5$ options.

$n$ is the starting value in the numerator.
$k$ is the number of factors (slots) in the denominator, counting down to $1$ .

Which of these answer choices represents the solution to the question below?

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

A. $(9 4)$
B. $(4 9)$
C. $(1 9)$
D. $(9 9)$
E. $(4 4)$

(spoiler)

Answer: B

This problem has already been solved earlier in this chapter. You only need to translate the setup into nCk notation.

The fraction below can be rewritten as 9C4 in nCk notation, as shown in answer B.

$\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$

In general, a question may look like a standard combination or permutation problem, but the answer choices will be written in nCk format. Solve the problem as usual, then choose the option that matches the correct notation.

Bringing it all together: question walkthrough video

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

Permutations

For any permutation or combination question, start by asking:

“How many decisions do I have to make, and how many options are there for each 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?

You’re making two decisions:

Choose a drink (5 options)
Choose a food item (10 options)

Write one blank for each decision, fill in the number of options, and multiply:

$\underline{5} \times \underline{10} = 50$

This same technique works for all permutation questions. The main challenge is figuring out how many options belong in each blank.

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

You’re filling 5 spots in a line, so you have 5 decisions.

First spot: 5 choices
Second spot: 4 choices (one child is already used)
Third spot: 3 choices
Fourth spot: 2 choices
Fifth spot: 1 choice

Multiply the choices for each spot:

$\underline{5} \times \underline{4} \times \underline{3} \times \underline{2} \times \underline{1} = 120$

Now 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.

First, count the number of ways to arrange the remaining 4 tasks if you ignore the shopping-before-grandma rule:

$\underline{4} \times \underline{3} \times \underline{2} \times \underline{1} = 24$

Now apply the constraint. In these 24 arrangements, grocery shopping and visiting grandma can appear in either order:

In half the arrangements, shopping comes before grandma.
In the other half, grandma comes before shopping.

So only 50% of the 24 arrangements are valid:

Valid permutations: $\frac{1}{2} \times 24 = 12$

Combinations

Order matters for permutations (for example, who is first in line versus second). Order 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?

A “pair” means you’re choosing 2 people, and it doesn’t matter who is chosen first.

Start the same way you would for a permutation:

First choice: 6 options
Second choice: 5 options

That gives $6 \times 5$ ordered selections. But each pair is counted twice (Alex then Blair, and Blair then Alex). To remove the duplicates, divide by $2!$ .

That exclamation mark means factorial. Here, $2! = 2 \times 1 = 2$ .

$\frac{6 \times 5}{2 \times 1} = 15$

So, $15$ pairs of news anchors can 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.

You’re choosing 4 coins, so start by counting ordered selections:

First coin: 9 options
Second coin: 8 options
Third coin: 7 options
Fourth coin: 6 options

$\underline{9} \times \underline{8} \times \underline{7} \times \underline{6} = 3024$

To remove duplicates, divide by “the-number-of-slots factorial.” There are 4 slots, so divide by $4!$ :

$\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

See how the number of slots in the divisor is the same as the number of slots in the numerator? They always match up for combination questions.

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

Equations for permutation and combination questions

The “blanks” method above is a practical way to apply the standard permutation and combination equations. If you prefer using formulas directly, use this section.

Permutations equation

Use permutations when the order of elements matters, like counting the number of ways people can line up.

Variables:

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

$Permutations = \frac{n !}{( n - k )!}$

Let’s solve the following question using this equation.

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

Which values match $n$ and $k$ ? Use them to create and solve the permutations equation.

(spoiler)

$n = 5$
$k = 5$

$Permutations = \frac{n !}{( n - k )!} = \frac{5 !}{( 5 - 5 )!} = 120$

There are 5 kindergarteners total, so $n = 5$ . You’re placing all 5 into the line, so $k = 5$ .

Here’s the full math equation:

$Permutations = \frac{n !}{( n - k )!} = \frac{5 !}{( 5 - 5 )!} = \frac{5 !}{0 !} = \frac{5 * 4 * 3 * 2 * 1}{1} = 120$

Sidenote

Zero factorial (0!) is equal to 1

It’s important to know that $0! = 1$ .

Think of it this way: there is only 1 way to group 0 things, meaning one group containing nothing.

Combinations equation

Use combinations when the order of elements does not matter, like counting the number of ways to pick marbles from a jar.

Variables:

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

$Combinations = \frac{n !}{k ! ( n - k )!}$

Let’s solve the following question using this equation.

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

Which values match $n$ and $k$ ? Use them to create and solve the combinations equation.

(spoiler)

$n = 6$
$k = 2$

$Combinations = \frac{n !}{k ! ( n - k )!} = \frac{6 !}{2 ! ( 6 - 2 )!} = 15$

There are 6 total news anchors to choose from, so $n = 6$ . A pair is a group of 2, so $k = 2$ .

Here’s the full math equation:

$Combinations = \frac{n !}{k ! ( n - k )!} = \frac{6 !}{2 ! ( 6 - 2 )!} = \frac{6 !}{2 ! * 4 !} = \frac{6 * 5 * 4 * 3 * 2 * 1}{( 2 * 1 ) * ( 4 * 3 * 2 * 1 )} = \frac{6 * 5}{2 * 1} = 15$

n Choose k notation (nCk)

Although it’s tested infrequently, you might see a notation that looks like a fraction without a dividing line. It represents the combinations calculation, which is why it uses “C.”

$(k n) = (3 5) = \frac{5 * 4 * 3}{3 * 2 * 1}$

The notation $(3 5)$ means “5C3” or “5 choose 3.” It’s another way to write the number of ways to choose a group of $3$ from $5$ options.

$n$ is the starting value in the numerator.
$k$ is the number of factors (slots) in the denominator, counting down to $1$ .

Which of these answer choices represents the solution to the question below?

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

A. $(9 4)$
B. $(4 9)$
C. $(1 9)$
D. $(9 9)$
E. $(4 4)$

(spoiler)

Answer: B

This problem has already been solved earlier in this chapter. You only need to translate the setup into nCk notation.

The fraction below can be rewritten as 9C4 in nCk notation, as shown in answer B.

$\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$

Bringing it all together: question walkthrough video

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