What is a combination?

A combination is a way to choose a group of items from a larger set where the order of selection does not matter. That last part is the key. If you pick Alex, Sam, and Jordan to form a 3-person team, it’s still the same team if you list them as Jordan, Alex, Sam. Combinations are the backbone of many real-world questions that sound like “How many different groups can I make?”

The classic notation is nCr, read as “n choose r” and written mathematically as \(\binom{n}{r}\). Here, n is the total number of distinct items available, and r is the size of the group you’re choosing. If you have 10 candidates and you want a committee of 3, the number of committees is \(\binom{10}{3}\).

The standard combinations formula is:
\(\binom{n}{r} = \frac{n!}{r!(n-r)!}\)

The exclamation mark means factorial. For any positive integer n, n! is the product of all integers from 1 to n. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials show up because counting “all possible orderings” is naturally a multiplication process.

Why does the combination formula look like a factorial fraction? Think of it in three steps:

• Step 1 — Count ordered selections: If order mattered, you’d be counting permutations. The number of ways to pick r items from n in order is \(n \times (n-1) \times \dots \times (n-r+1)\). That’s \( \frac{n!}{(n-r)!} \).
• Step 2 — Remove “internal re-ordering” duplicates: Each group of r items can be arranged in r! different orders, but all those orders represent the same combination. Divide by r!.
• Step 3 — You get combinations: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).

Sometimes you can pick the same “type” more than once. For example, if a donut shop has 6 flavors and you are buying 3 donuts, you might choose 2 chocolate and 1 glazed. That’s a combinations-with-repetition problem. The formula becomes:
\(\binom{n+r-1}{r}\)

Intuition: you’re still forming a group of size r, but duplicates are allowed, so the counting space is larger. This calculator supports both modes so you can handle standard “choose a team” scenarios and “choose with repeats” scenarios.

• Symmetry: \(\binom{n}{r} = \binom{n}{n-r}\). Choosing r items to keep is the same as choosing n−r items to leave out.
• Edge values: \(\binom{n}{0} = 1\) and \(\binom{n}{n} = 1\). There’s exactly one way to choose nothing or everything.
• Always an integer: Even though it’s a fraction of factorials, the result is always a whole number.
• Binomial connection: The coefficients in \((a+b)^n\) are combinations: \((a+b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r\).

Worked examples you can screenshot

You have 10 people and you want a team of 3. Order doesn’t matter, so use combinations:
\(\binom{10}{3} = \frac{10!}{3!7!} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120\)

A standard 5-card poker hand is any group of 5 cards from a 52-card deck. Order doesn’t matter:
\(\binom{52}{5} = 2,598,960\)

That number is why poker has so much variety. Two million+ possible hands means you can play a lot of games without repeating the same exact set of cards.

Choose 2 toppings from 9 options:
\(\binom{9}{2} = \frac{9 \cdot 8}{2 \cdot 1} = 36\)

Choose 3 donuts from 6 flavors (you can repeat flavors). That’s combinations with repetition:
\(\binom{6+3-1}{3} = \binom{8}{3} = 56\)

• If r = 1, then \(\binom{n}{1} = n\). Choosing 1 item from n gives n possible choices.
• If r = 2, then \(\binom{n}{2} = \frac{n(n-1)}{2}\). Great for “how many pairs?” questions.
• If r is near n, use symmetry to compute faster: \(\binom{100}{97} = \binom{100}{3}\).

How this calculator computes nCr (without overflowing)

Many combination calculators compute factorials directly (n!, r!, (n-r)!) and then divide. That approach is fine for small inputs, but factorials explode extremely fast. Even \(100!\) is far bigger than what normal floating-point numbers can represent exactly. If you’ve ever seen “Infinity” or weird rounding, that’s why.

This calculator uses exact integer arithmetic (JavaScript BigInt) and a multiplicative reduction method. Instead of computing n! and dividing huge numbers, it computes:

• \(\binom{n}{r} = \prod_{i=1}^{k} \frac{n-k+i}{i}\) where \(k = \min(r, n-r)\)
• At each step, it simplifies fractions using a greatest common divisor (GCD) so intermediate values stay manageable.
• The final result is exact — an integer with no rounding.

If you enable Show steps, you’ll also see a human-readable breakdown of the formula used and the main numbers involved. That’s helpful for learning or writing solutions by hand.

Use “with repetition” when your problem allows choosing the same kind of item multiple times, like selecting scoops of ice cream, distributing identical tokens into bins, or counting multisets. The math becomes \(\binom{n+r-1}{r}\) and is still computed exactly.

Huge values like \(\binom{1000}{500}\) are valid, but they produce extremely long integers (hundreds of digits). That’s not a “bug” — it’s what the math actually is. For sharing, the calculator also shows a compact scientific-notation preview.

Frequently Asked Questions

• What does nCr mean?

  nCr means “n choose r.” It counts how many different groups of size r you can choose from n distinct items when the order of the group does not matter.

• How is a combination different from a permutation?

  In a combination, order doesn’t matter: {A, B, C} is the same as {C, B, A}. In a permutation, order matters: ABC and CBA are different. If your problem involves arranging or ordering, you want permutations (nPr).

• Why is the combination formula divided by r!?

  Because when you count ordered selections (permutations), each unique group of r items is counted multiple times — once for every possible ordering. There are r! ways to rearrange r items, so dividing by r! removes those duplicates.

• Can nCr be a decimal?

  No. nCr is always a whole number because it counts a number of discrete groups. Even though the formula uses division, the final result is guaranteed to be an integer.

• What does “with repetition” mean?

  “With repetition” means you can choose the same kind of item more than once. For example, choosing 3 candies from 5 flavors where you can pick multiple candies of the same flavor. The formula becomes \(\binom{n+r-1}{r}\).

• What if r is bigger than n?

  In standard combinations (no repetition), choosing more items than exist doesn’t make sense, so the result is undefined. This calculator will prompt you to correct the inputs. In “with repetition” mode, r can be larger than n because repeats are allowed.

• Does this work for big numbers like 200 choose 100?

  Yes. The calculator uses BigInt and simplification to keep results exact. Note that the output can be very long — that’s expected. For convenience, you’ll also get a compact scientific-notation preview.