# Permutations

> A permutation is one way of picking some items and lining them up in a set order, and this formula counts how many such lineups exist.

Page: https://formulabro.yunusemrevurgun.com/computer-science/permutations/  
Category: Computer Science  
Published: 2026-10-22  
Updated: 2026-10-22

## Formula

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

Read it out loud: "P of n, k, equals n factorial, divided by n minus k factorial. P is the number of permutations, n is how many different items there are, and k is how many of them you pick and line up in order. Factorial means multiply a whole number by every whole number below it, down to 1."

## Before you start

- **Factorial, $n!$**: $n!$ is read 'n factorial'. It means multiply $n$ by every whole number below it, down to 1. For example $4! = 4 \times 3 \times 2 \times 1 = 24$, and $5! = 120$. Factorials grow very fast: $10! = 3,628,800$. By a rule everyone agrees on, $0! = 1$, which keeps formulas like this one working.
- **Powers, like $10^{4}$**: $10^{4}$ is read 'ten to the power 4'. The small raised number says how many times to multiply the number by itself, so $10^{4} = 10 \times 10 \times 10 \times 10 = 10000$. In the same way $3^{2} = 9$. Powers come up when each spot has the same number of choices, as in a PIN code.
- **Order matters**: Two lineups count as different when they put the same items in a different order. If runners A, B and C finish in that order, that is a different result from B, A, C, even though the same three runners are involved. A **permutation** counts these ordered lineups, while a **combination** ignores the order.
- **Multiplying counts of choices**: When a choice is made in several steps, the number of ways is the number of options at each step multiplied together. If a shirt comes in 3 colors and 2 sizes, there are $3 \times 2 = 6$ shirts in total. This multiplying of choices is the main idea behind every permutation count.
- **Subtracting, like $n - k$**: $n - k$ is the number of items left over after $k$ of them are chosen. If there are 8 runners and 3 places on the podium, then $8 - 3 = 5$ runners are not on the podium. This leftover count appears in the bottom of the formula.
- **No repeats**: The permutation formula assumes that each item can be used only once. A choice can repeat when the same item may be used again, such as a digit in a PIN, where the digit 7 can appear twice. When repeats are allowed, the counting rule is different, as the mistakes section explains.

More notation help: https://formulabro.yunusemrevurgun.com/math-basics/

## What each symbol means

- $P(n, k)$: The **number of permutations**: how many different ordered ways there are to choose and line up $k$ items out of $n$ different items. It is a count of arrangements, so it has no unit. For $n = 8$ and $k = 3$, it equals 336.
- $n$: The **number of different items** available to choose from, a whole number such as 8 runners. It counts the items in the full pool, so it has no unit. It is always 0 or more.
- $k$: The **number of spots** you fill in order, a whole number such as 3 places on a podium. When items cannot repeat, $k$ cannot be bigger than $n$, because there would not be enough items to fill the spots.
- $n!$: The **factorial** of $n$, the product of every whole number from $n$ down to 1. For example $8! = 40320$. It counts the ways to line up all $n$ items in a single row, so it has no unit.
- $(n-k)!$: The **factorial of the items left over**, which is $n - k$. For 8 runners and 3 spots, $(8 - 3)! = 5! = 120$. Dividing by it cancels the extra orderings of the items that are not in the chosen spots.

## The formula piece by piece

- $n$: The number of different items available. In the podium example, $n = 8$ runners. Each spot can be filled by any runner who is still available, and the number of available runners goes down as spots are filled.
- $k$: The number of spots you fill, in order. For a podium with gold, silver and bronze, $k = 3$. It is a whole number from 0 up to $n$, because with no repeats you cannot fill more spots than there are items.
- $n-k$: The number of items left over after the $k$ spots are filled. For 8 runners and 3 spots, that is $8 - 3 = 5$ runners who are not on the podium. The bottom of the fraction uses this leftover count.
- $n!$: The number of ways to line up all $n$ items in a single row. For 8 runners, $8! = 40320$ ways. It is the count before any spots are cut off, because every item is used in every position.
- $(n-k)!$: The number of ways to line up the leftover items. When $n!$ is divided by $(n-k)!$, the factors from $n - k$ down to 1 cancel, and only the factors for the $k$ filled spots remain.
- $P(n, k) = \frac{n!}{(n-k)!}$: In one plain sentence: the number of ordered ways to choose and line up $k$ items from $n$ different items, with no repeats, equals $n$ factorial divided by the factorial of $n$ minus $k$.

## What the formula is doing, and why

**What goes in and what comes out.** You put in two whole numbers: $n$, the number of different items, and $k$, the number of spots to fill. You get out a count of the ordered ways to fill the spots without repeats. Order matters, so the lineup $ABC$ is different from $BCA$. The answer is a whole number with no unit.

**Why multiply the choices?** Fill the spots one at a time. The first spot can take any of the $n$ items. The second spot can take any of the other $n - 1$ items, because the first item is already used. The third spot has $n - 2$ choices, and the $k$th spot has $n - k + 1$ choices. Multiply all the choices together to get $n(n-1)(n-2)\cdots(n-k+1)$.

**Why the factorial form?** $n!$ is the product of all $n$ numbers. Dividing by $(n-k)!$ removes the factors from $n - k$ down to 1, which leaves exactly the $k$ factors from $n$ down to $n - k + 1$. Those are the choices we multiplied. Two edge cases check the formula. When $k = n$, every item is used, and the answer is $n!$. When $k = 0$, nothing is picked, and the answer is 1, the single empty lineup.

**When does it not apply?** The formula is only for items that cannot repeat. If repeats are allowed, such as the digits in a PIN, every spot has all $n$ choices again, and the answer is $n^{k}$. Also, $k$ cannot be bigger than $n$ when there are no repeats, because there would not be enough items to fill the spots.

## An everyday picture

Think of a playlist. You have 10 songs and want to play 3 of them, in order. The first slot has 10 choices, the second has 9 songs left, and the third has 8 left. The playlist that plays A, B, C is different from the one that plays B, A, C, even though both use the same songs. So the count is $10 \cdot 9 \cdot 8 = 720$ playlists, which is $P(10, 3)$.

## Worked example: Podium places in an 8-runner race

Eight runners race. The top three finishers take gold, silver and bronze. Those are different places, so the order of the three names matters. Here $n = 8$ runners and $k = 3$ podium places. We want $P(8, 3)$, the number of different ways to fill the three places.

1. **Write the formula and put in the numbers.** The formula says $P(n, k) = \frac{n!}{(n-k)!}$. With $n = 8$ and $k = 3$, the bottom becomes $(8 - 3)! = 5!$, so $P(8, 3) = \frac{8!}{5!}$.
2. **Compute the factorials.** $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$. And $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. Both start at a top number and multiply down to 1, so the bottom factorial is part of the top one.
3. **Divide.** $\frac{40320}{120} = 336$. The division removes the $5!$ part of $8!$, so all that is left is $8 \times 7 \times 6$, one factor for each of the three spots.
4. **Check by filling the spots.** Gold can go to any of the 8 runners. Silver can go to any of the 7 runners who are left. Bronze can go to any of the 6 runners who are still left. So $8 \times 7 \times 6 = 336$, which matches the formula.
5. **Cross-check with groups.** Ignoring order, there are 56 different groups of 3 runners. Each group can be arranged in $3! = 6$ orders, such as gold, silver, bronze in six ways. So $56 \times 6 = 336$, which again gives 336 different podium results.

**Answer:** There are **336** different podium results from 8 runners. Each different order of the top three is a different result.

## Common mistakes

- Ignoring order. Counting the podium as a set of three runners gives a smaller number, because gold, silver, bronze and bronze, silver, gold are different results. When order does not matter, use combinations instead.
- Multiplying $n$ by $k$. $P(8, 3)$ is $8 \times 7 \times 6 = 336$, not $8 \times 3 = 24$. Count down from $n$ for $k$ spots, one spot at a time.
- Using this formula when items can repeat. The digits in a PIN can repeat, so the right count is $10^{4} = 10000$ options, not $P(10, 4) = 5040$.
- Asking for more items than exist. With 3 items and no repeats, there is no way to line up 5 of them, so $P(3, 5)$ does not exist. Make sure $k \le n$.

## Where you will meet it

- Race results and rankings, where first, second and third place are different outcomes.
- Counting anagrams, which are rearrangements of letters. The word CAT has $3! = 6$ arrangements of its three different letters: CAT, CTA, ACT, ATC, TAC and TCA.
- Brute-force search in computing, which means trying every possibility one by one. A program can try every order of a list of distinct stops on a delivery route to find the shortest route.
- Counting codes or passwords that cannot repeat a character, where the first character has $n$ choices and the next has one fewer.

## Questions people ask

### What is the difference between a permutation and a combination?

In a permutation the order matters, so the lineup ABC is different from BCA. In a combination the order does not matter, so ABC and BCA are the same group. Each group of $k$ items can be lined up in $k!$ orders, so the number of combinations is $\frac{P(n, k)}{k!}$.

### Why is $0! = 1$?

It is a rule that keeps formulas like this one working. With $k = n$, the formula gives $P(n, n) = \frac{n!}{0!}$. That must equal $n!$, because every item is used. The only value of $0!$ that makes this true is 1.

### How many ways can I arrange all $n$ items?

There are $n!$ ways. Five books on a shelf can be arranged in $5! = 120$ orders. That is the same as $P(5, 5)$, because all five books are used and the shelf is filled in every possible order.

## Related formulas

- [Big O Notation](https://formulabro.yunusemrevurgun.com/computer-science/big-o-notation/)
- [Bits Needed](https://formulabro.yunusemrevurgun.com/computer-science/bits-needed/)
- [Sum of the First n Integers](https://formulabro.yunusemrevurgun.com/computer-science/sum-of-first-n-integers/)

## Sources

- [Permutation, Wikipedia](https://en.wikipedia.org/wiki/Permutation)
- [Factorial, Wikipedia](https://en.wikipedia.org/wiki/Factorial)

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