What's the difference between combinations and permutations?

Combinations count unordered selections—choosing (A, B) and (B, A) counts as the same selection. Permutations count ordered arrangements—(A, B) and (B, A) are different. Use combinations when selecting a group (like a committee) and permutations when arranging items in a specific order (like password digits).

Can you have a combination where r is bigger than n?

No. You cannot choose more items than you have available. If r > n, then \binom{n}{r} = 0. Always check that r \leq n before calculating.

Why do we divide by r! in the combination formula?

We divide by r! because permutations count each different arrangement separately, but combinations treat all arrangements of the same items as identical. Dividing by r! removes this overcounting by accounting for the number of ways to arrange r items among themselves.

Is C(n, r) the same as C(n, n - r)?

Yes. \binom{n}{r} = \binom{n}{n-r} always holds. This is because choosing r items to include is the same as choosing n - r items to exclude. You can use this to simplify calculations when r is large—for example, \binom{100}{97} = \binom{100}{3} is much easier to calculate.

Combinations

Combinations count the ways to select items from a group when order doesn't matter; they are used for choosing unordered subsets.

\binom{n}{r} = \frac{n!}{r!(n-r)!}

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What each symbol means

What Combinations takes

Symbol	Meaning
$n$	The total number of items to choose from; must be a non-negative integer.
$r$	The number of items you are choosing; must be a non-negative integer with $r \leq n$ .

When to use it

When you need to count selections where different arrangements of the same items count as one—like choosing committee members or lottery numbers.

Level

Usually taught in: Algebra II · Appears on: SAT, ACT

Worked examples

1. Simple combination of small integers

Problem

Evaluate

\binom{4}{2}

$\binom{4}{2} = \frac{4!}{2!(4-2)!}$
Write out the combination formula with $n = 4$ and $r = 2$ .
$= \frac{4!}{2! \cdot 2!}$
Simplify the denominator: $4 - 2 = 2$ .
$= \frac{24}{2 \cdot 2}$
Calculate the factorials: $4! = 24$ and $2! = 2$ .
$= \frac{24}{4}$
Multiply the denominator: $2 \cdot 2 = 4$ .
$= 6$
Divide: $24 \div 4 = 6$ .

Answer: $\binom{4}{2} = 6$

This shows that there are 6 ways to choose 2 items from 4 items when order doesn't matter. Direct calculation is straightforward with small numbers.

2. Combination with larger numbers and factorization

Problem

How many ways can 4 people be chosen from a group of 10?

$\binom{10}{4} = \frac{10!}{4!(10-4)!}$
Write the combination formula with $n = 10$ and $r = 4$ .
$= \frac{10!}{4! \cdot 6!}$
Simplify: $10 - 4 = 6$ .
$= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6!}{4! \cdot 6!}$
Expand $10! = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6!$ to show cancellation.
$= \frac{10 \cdot 9 \cdot 8 \cdot 7}{4!}$
Cancel the $6!$ terms; the key is not computing $10!$ fully.
$= \frac{10 \cdot 9 \cdot 8 \cdot 7}{24}$
Calculate $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$ .
$= \frac{5040}{24}$
Multiply the numerator: $10 \cdot 9 = 90$ , $90 \cdot 8 = 720$ , $720 \cdot 7 = 5040$ .
$= 210$
Divide: $5040 \div 24 = 210$ .

Answer: $\binom{10}{4} = 210$

There are 210 ways to choose 4 people from a group of 10. This example shows the importance of canceling factorials before multiplying—otherwise you'd be working with huge numbers unnecessarily.

3. Real-world selection problem

Problem

A lottery requires you to choose 6 different numbers from 1 to 50. How many different lottery tickets are possible?

$\binom{50}{6} = \frac{50!}{6!(50-6)!}$
This is a combination because the order you select the numbers doesn't matter—ticket $(1,2,3,4,5,6)$ is identical to $(6,5,4,3,2,1)$ .
$= \frac{50!}{6! \cdot 44!}$
Simplify: $50 - 6 = 44$ .
$= \frac{50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45}{6!}$
Write out the numerator by canceling $44!$ ; calculate $6! = 720$ .
$= \frac{11,441,304,000}{720}$
Multiply the numerator: $50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 = 11,441,304,000$ .
$= 15,890,700$
Divide: $11,441,304,000 \div 720 = 15,890,700$ .

Answer: $\binom{50}{6} = 15,890,700$

There are nearly 16 million possible lottery tickets when choosing 6 numbers from 50. This real-world example shows why combinations matter: it would be impossible to list every possibility by hand, and the formula gives us an exact count instantly.

Common mistakes

Where Combinations usually goes wrong

The mistake	Why it is wrong	The fix
Calculating $\binom{5}{3}$ as $5 \cdot 4 \cdot 3 = 60$ .	This is the permutation formula, which counts ordered arrangements; combinations must divide by $r!$ to eliminate duplicate selections.	Use the full formula: $\binom{5}{3} = \frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1} = \frac{60}{6} = 10$ .
Trying to compute $\binom{5}{8}$ or another combination where $r > n$ .	You cannot choose 8 items from only 5 available items, so this violates the definition of combinations and is mathematically impossible.	Always verify that $r \leq n$ before calculating; if $r > n$ , the answer is 0.
Flipping $n$ and $r$ , writing $\binom{3}{10}$ instead of $\binom{10}{3}$ for a problem choosing 3 from 10.	It is easy to confuse which number is the total pool and which is the selection size, especially when reading a word problem quickly.	Remember that $n$ is the total available items and $r$ is the number you select; reread the problem if unsure.

Tips and when to use something else

Use $\binom{n}{r} = \binom{n}{n-r}$ as a shortcut: choosing 3 items to keep from 10 is the same as choosing 7 items to exclude.
Combinations and permutations are related: $\binom{n}{r} = \frac{P(n,r)}{r!}$ , so use Permutations when order matters.
For large values of $n$ and $r$ , always cancel factorial terms first before multiplying to avoid working with enormous numbers.
The Fundamental Counting Principle helps you decide whether you need combinations (unordered groups) or permutations (ordered arrangements).

Related formulas

PermutationsPermutations count the number of ways to arrange a set of objects in a specific order, where changing the order creates a different arrangement.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.Binomial ProbabilityBinomial Probability finds the likelihood of exactly k successes in n independent identical trials, each with fixed probability p.Basic ProbabilityBasic Probability calculates the likelihood of an event by dividing favorable outcomes by the total number of possible outcomes.

Frequently asked questions

What's the difference between combinations and permutations?: Combinations count unordered selections—choosing $(A, B)$ and $(B, A)$ counts as the same selection. Permutations count ordered arrangements— $(A, B)$ and $(B, A)$ are different. Use combinations when selecting a group (like a committee) and permutations when arranging items in a specific order (like password digits).
Can you have a combination where r is bigger than n?: No. You cannot choose more items than you have available. If $r > n$ , then $\binom{n}{r} = 0$ . Always check that $r \leq n$ before calculating.
Why do we divide by r! in the combination formula?: We divide by $r!$ because permutations count each different arrangement separately, but combinations treat all arrangements of the same items as identical. Dividing by $r!$ removes this overcounting by accounting for the number of ways to arrange $r$ items among themselves.
Is C(n, r) the same as C(n, n - r)?: Yes. $\binom{n}{r} = \binom{n}{n-r}$ always holds. This is because choosing $r$ items to include is the same as choosing $n - r$ items to exclude. You can use this to simplify calculations when $r$ is large—for example, $\binom{100}{97} = \binom{100}{3}$ is much easier to calculate.

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Reviewed 2026-09-18