What is the difference between permutations and combinations?

Permutations count arrangements where order matters; combinations count selections where order does not. If you're arranging people in seats, use permutations. If you're choosing a team, use combinations. The permutation formula is P(n, r) = \frac{n!}{(n-r)!} and the combination formula is C(n, r) = \frac{n!}{r!(n-r)!}.

When should I use P(n, r) instead of the multiplication principle?

Both methods work, but the permutation formula is faster when you want a single calculation. Use the multiplication principle (count choices at each step) when you want to understand the structure of the problem or when the counting is unequal at each step. For equal decreasing choices, P(n, r) gives the answer directly.

Can you compute P(n, 0)?

P(n, 0) = \frac{n!}{n!} = 1 by definition, because there is exactly one way to arrange zero objects from n objects: the empty arrangement. This makes sense and is consistent across counting formulas.

Why is (n-r)! in the denominator and not just (n-r)?

The (n-r)! cancels out the trailing factors in n! that you do not need. When you expand n!, you get n \times (n-1) \times \cdots \times 2 \times 1, and (n-r)! = (n-r) \times (n-r-1) \times \cdots \times 2 \times 1. Dividing removes those factors, leaving you with exactly r factors: n \times (n-1) \times \cdots \times (n-r+1).

Permutations

Permutations count the number of ways to arrange a set of objects in a specific order, where changing the order creates a different arrangement.

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

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What each symbol means

What Permutations takes

Symbol	Meaning
$n$	The total number of objects available to choose from; must be a non-negative integer, and if treated as the number of items to arrange rather than the total available, the result will be incorrect.
$r$	The number of objects you are selecting and arranging; must be a non-negative integer not exceeding $n$ , and if $r > n$ , the formula is undefined because you cannot arrange more items than you have available.

When to use it

Use permutations when you need to count arrangements where order matters and you're selecting from a larger set.

Level

Usually taught in: Algebra II · Appears on: ACT

Worked examples

1. Arranging books on a shelf

Problem

How many ways can you arrange 3 books on a shelf if you have 5 books to choose from?

$P(5, 3) = \frac{5!}{(5-3)!}$
We substitute $n = 5$ and $r = 3$ into the permutation formula.
$= \frac{5!}{2!}$
We simplify the denominator by computing $5 - 3 = 2$ .
$= \frac{120}{2}$
We evaluate the factorials: $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ and $2! = 2 \times 1 = 2$ .
$= 60$
We divide $120$ by $2$ to get the final answer.

Answer: $60$

We use the permutation formula because we're selecting 3 books from 5 and the order in which we arrange them matters (book A then B then C is different from C then B then A).

2. Arranging relay runners

Problem

In how many ways can a coach select and order 4 runners from a team of 10 for a relay race?

$P(10, 4) = \frac{10!}{(10-4)!}$
Substitute $n = 10$ and $r = 4$ into the permutation formula.
$= \frac{10!}{6!}$
Simplify the denominator: $10 - 4 = 6$ .
$= \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!}$
Expand $10!$ to show the factors up to and including $6!$ , which allows us to see the cancellation.
$= 10 \times 9 \times 8 \times 7$
Cancel the $6!$ from numerator and denominator.
$= 5040$
Multiply: $10 \times 9 = 90$ , $90 \times 8 = 720$ , and $720 \times 7 = 5040$ .

Answer: $5040$

This problem shows the cancellation technique: rather than computing $10!$ and $6!$ separately, we can cancel $(n-r)!$ from the numerator and denominator, leaving only the product of the first $r$ terms. This is especially useful with large factorials.

3. Selecting and ordering time points for temperature measurement

Problem

A laboratory is conducting an experiment where researchers measure the temperature at 7 different times throughout a day. They need to select 4 of these time points and arrange them in order of measurement intensity (from lowest to highest reading priority). In how many different orders can they arrange 4 time points chosen from the 7?

$P(7, 4) = \frac{7!}{(7-4)!}$
Identify that $n = 7$ total time points and $r = 4$ selected time points to arrange.
$= \frac{7!}{3!}$
Simplify: $7 - 4 = 3$ .
$= 7 \times 6 \times 5 \times 4$
Cancel the $3!$ from numerator and denominator, leaving the product of exactly $r = 4$ consecutive descending integers starting from 7.
$= 840$
Calculate: $7 \times 6 = 42$ , $42 \times 5 = 210$ , and $210 \times 4 = 840$ .

Answer: $840$

In this lab scenario, order matters because the researchers care about which specific time points come first, second, third, and fourth. Permutations count all these distinct orderings, which is why we use the permutation formula rather than combinations.

Common mistakes

Where Permutations usually goes wrong

The mistake	Why it is wrong	The fix
Trying to compute $P(5, 7)$ or writing $P(5, 7) = \frac{5!}{-2!}$ when selecting more items than available.	When $r > n$ , the quantity $(n - r)$ becomes negative, making $(n-r)!$ undefined because factorials of negative numbers do not exist.	Always check that $r \leq n$ before applying the formula. If $r > n$ , there are zero permutations because you cannot arrange more objects than you have.
Using the formula $P(n, r) = \frac{n!}{r!(n-r)!}$ (which is actually the combination formula) instead of $P(n, r) = \frac{n!}{(n-r)!}$ .	The combination formula divides by $r!$ to account for the fact that order does not matter; permutations do not divide by $r!$ because order does matter and creates distinct arrangements.	Use $P(n, r) = \frac{n!}{(n-r)!}$ for permutations. Only add an $r!$ in the denominator if the problem explicitly states that order does not matter, in which case use combinations instead.
Writing a product with the wrong number of factors, such as $P(5,3) = 5 \times 4$ instead of $5 \times 4 \times 3$ .	The product must contain exactly $r$ factors (the number of objects being arranged). Writing too few factors gives a result that is too small.	The product form is $n \times (n-1) \times (n-2) \times \cdots \times (n-r+1)$ , which has exactly $r$ terms. For $P(5,3)$ , write $5 \times 4 \times 3$ (three factors), not $5 \times 4$ (two factors).

Tips and when to use something else

Use Combinations instead of Permutations when order does not matter — for example, choosing a committee where Bob-Alice-Carol is the same as Alice-Bob-Carol.
The product form $P(n, r) = n \times (n-1) \times (n-2) \times \cdots \times (n-r+1)$ is often faster than computing factorials by hand, especially for large $n$ .
Permutations appear in real-world problems like arranging people in a line, ordering tasks, ranking winners, or passwords where repetition is not allowed.
If the problem allows repetition (such as a 4-digit PIN where you can reuse digits), use the Fundamental Counting Principle instead: multiply $n \times n \times n \times n = n^r$ .

Related formulas

CombinationsCombinations count the ways to select items from a group when order doesn't matter; they are used for choosing unordered subsets.Fundamental Counting PrincipleMultiply the number of choices at each step to find the total number of different possible outcomes when making sequential selections or decisions.Multiplication RuleFind the probability that two events both occur when one might affect the chance of the other using multiplication of probabilities.

Frequently asked questions

What is the difference between permutations and combinations?: Permutations count arrangements where order matters; combinations count selections where order does not. If you're arranging people in seats, use permutations. If you're choosing a team, use combinations. The permutation formula is $P(n, r) = \frac{n!}{(n-r)!}$ and the combination formula is $C(n, r) = \frac{n!}{r!(n-r)!}$ .
When should I use P(n, r) instead of the multiplication principle?: Both methods work, but the permutation formula is faster when you want a single calculation. Use the multiplication principle (count choices at each step) when you want to understand the structure of the problem or when the counting is unequal at each step. For equal decreasing choices, $P(n, r)$ gives the answer directly.
Can you compute P(n, 0)?: $P(n, 0) = \frac{n!}{n!} = 1$ by definition, because there is exactly one way to arrange zero objects from $n$ objects: the empty arrangement. This makes sense and is consistent across counting formulas.
Why is (n-r)! in the denominator and not just (n-r)?: The $(n-r)!$ cancels out the trailing factors in $n!$ that you do not need. When you expand $n!$ , you get $n \times (n-1) \times \cdots \times 2 \times 1$ , and $(n-r)! = (n-r) \times (n-r-1) \times \cdots \times 2 \times 1$ . Dividing removes those factors, leaving you with exactly $r$ factors: $n \times (n-1) \times \cdots \times (n-r+1)$ .

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Reviewed 2026-09-18