Combinations Calculator

What is a combinations calculator?

A combinations calculator works out how many different ways you can choose a group of items from a larger set when the order of selection does not matter. This quantity is called the number of combinations, written as ${}^{n}C_{r}$, “n choose r”, or with the binomial coefficient $\binom{n}{r}$. Here $n$ is the total number of available items and $r$ is how many of them you pick.

Combinations appear whenever you care only about which items end up together, not the sequence in which they were chosen. Picking 2 toppings out of 5 gives the same pizza no matter which topping you name first, so it is a combination problem. If the order mattered, you would be counting permutations instead.

How does it work?

Enter the total number of items $n$ and the number you want to choose $r$, and the calculator returns ${}^{n}C_{r}$ instantly. Both values must be whole numbers, and $r$ cannot be larger than $n$ — you cannot choose more items than you have. If $r > n$, or either field is left empty, the result stays blank.

Formula

The number of combinations is given by the binomial coefficient:

${}^{n}C_{r} = \binom{n}{r} = \frac{n!}{r!\,(n-r)!}$

Here $n!$ (n factorial) is the product of all positive integers up to $n$, so $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. By convention $0! = 1$, which is why choosing zero items, or choosing all of them, always gives exactly one combination.

A few useful identities follow directly from the formula:

1. $\binom{n}{0} = 1$ — there is one way to choose nothing.
2. $\binom{n}{n} = 1$ — there is one way to choose everything.
3. $\binom{n}{r} = \binom{n}{n-r}$ — choosing $r$ to keep is the same as choosing $n-r$ to leave out.

Worked examples

1. Example 1: Choose 2 items from 5. $\binom{5}{2} = \frac{5!}{2!\,3!} = \frac{120}{2 \times 6} = 10$.
2. Example 2: Choose 3 items from 10. $\binom{10}{3} = \frac{10!}{3!\,7!} = \frac{3628800}{6 \times 5040} = 120$.
3. Example 3: Choose all 5 from 5. $\binom{5}{5} = \frac{5!}{5!\,0!} = 1$.
4. Example 4: Choose 0 from 5. $\binom{5}{0} = \frac{5!}{0!\,5!} = 1$.

Practical notes

• Combinations count unordered selections. If the arrangement matters — for example seating people in a row — use permutations, where ${}^{n}P_{r} = \frac{n!}{(n-r)!}$.
• The values grow quickly because of the factorials, so even modest inputs can yield very large counts.
• Combinations underpin probability, the binomial distribution, lottery odds, card-hand counting, and combinatorial design problems.

Frequently asked questions

What is the difference between combinations and permutations?

In combinations the order of the chosen items does not matter, so $\{A, B\}$ and $\{B, A\}$ count as one selection. In permutations order matters, so they count as two. As a result there are always at least as many permutations as combinations for the same $n$ and $r$.

Why is choosing 0 items equal to 1?

Because $0! = 1$, the formula gives $\binom{n}{0} = \frac{n!}{0!\,n!} = 1$. Intuitively, there is exactly one way to select nothing at all — the empty selection.

Can r be larger than n?

No. You cannot choose more items than exist in the set, so $\binom{n}{r}$ is only defined for $0 \le r \le n$. This calculator returns a blank result when $r > n$.