Combination Calculator – Calculate nCr Online

Calculate combinations (nCr) instantly with our free online combination calculator. Find how many ways r items can be chosen from n items using the combination formula.

n (total items)

r (items to choose)

Examples:

Understanding Combinations

Combinations count the number of ways you can select items from a larger set when the order doesn't matter. If you're picking 3 toppings for your pizza from 10 available options, it doesn't matter whether you pick pepperoni first or last – you end up with the same pizza. That's a combination.

This is different from permutations, where order does matter. A lock combination is actually a permutation – 1-2-3 opens the lock, but 3-2-1 does not. In true combinations, selecting Alice, Bob, and Carol for a committee is the same as selecting Carol, Alice, and Bob.

Combinations appear everywhere: lottery drawings, card hands, committee selections, menu combinations, and sample selections in research. Anytime you're choosing a subset without regard to order, you're dealing with combinations.

The Combination Formula

C(n,r) = n! / (r! × (n-r)!)

Also written as _nC_r, C_n,r, or the binomial coefficient (ⁿ⁄_r)

Total number of items available

Number of items to choose

Factorial: multiply all integers from 1 to that number

Key Properties

C(n,0) = 1

There's one way to choose nothing

C(n,n) = 1

There's one way to choose everything

C(n,r) = C(n, n-r)

Choosing r to include = choosing (n-r) to exclude

C(n,1) = n

There are n ways to choose 1 item from n

Combinations vs Permutations

Combinations (order doesn't matter)

C(n,r) = P(n,r) / r!

Pizza toppings, lottery numbers, committee members

Permutations (order matters)

P(n,r) = n! / (n-r)!

Password codes, race rankings, seating arrangements

Worked Examples

Example 1: Poker Hands

How many different 5-card hands can be dealt from a standard 52-card deck?

n = 52 (total cards), r = 5 (cards in hand)

C(52,5) = 52! / (5! × 47!)

= (52 × 51 × 50 × 49 × 48) / (5 × 4 × 3 × 2 × 1)

= 311,875,200 / 120

= 2,598,960 possible hands

This is why getting a specific hand like a royal flush is so rare – there's only 4 ways out of nearly 2.6 million possibilities.

Example 2: Committee Selection

A club has 12 members. How many ways can they choose a 4-person committee?

n = 12 (members), r = 4 (committee size)

C(12,4) = 12! / (4! × 8!)

= (12 × 11 × 10 × 9) / (4 × 3 × 2 × 1)

= 11,880 / 24

= 495 possible committees

Notice we only multiply the top 4 terms of 12! because the 8! cancels with the denominator's 8!.

Example 3: Pizza Toppings

A pizza shop offers 8 toppings. How many different 3-topping pizzas can you order?

n = 8 (toppings), r = 3 (toppings per pizza)

C(8,3) = 8! / (3! × 5!)

= (8 × 7 × 6) / (3 × 2 × 1)

= 336 / 6

= 56 different pizzas

If you could also choose 0, 1, 2, 4, 5, 6, 7, or 8 toppings, the total would be C(8,0) + C(8,1) + ... + C(8,8) = 2⁸ = 256 possible pizzas.

Example 4: Lottery Odds

A lottery requires choosing 6 numbers from 1 to 49. What are your odds of winning?

n = 49 (numbers), r = 6 (choices)

C(49,6) = 49! / (6! × 43!)

= (49 × 48 × 47 × 46 × 45 × 44) / (6 × 5 × 4 × 3 × 2 × 1)

= 10,068,347,520 / 720

= 13,983,816 possible combinations

Your odds of winning are 1 in 13,983,816. This is why lotteries can offer such large jackpots – the probability of any single ticket winning is extremely low.

Quick Fact

Blaise Pascal (1623-1662) studied combinations extensively and arranged them in what we now call Pascal's Triangle. Each entry in the triangle is a combination value C(n,r). The triangle reveals patterns in binomial expansions, probability, and even fractal geometry (the Sierpinski triangle emerges when you color odd numbers). Pascal's work on combinations with Pierre de Fermat also founded probability theory – they developed it while solving a gambler's problem about how to fairly divide stakes in an interrupted game.

Frequently Asked Questions

When do I use combinations instead of permutations?

Use combinations when order doesn't matter – selecting team members, choosing menu items, drawing cards. Use permutations when order matters – creating passwords, ranking finishers in a race, arranging books on a shelf. Ask yourself: if I rearrange my selections, is it still the same outcome? If yes, it's a combination.

Why does C(n,r) equal C(n, n-r)?

Choosing r items to include automatically determines which (n-r) items you're excluding. If you have 10 people and choose 3 for a committee, you're also choosing 7 people who won't be on the committee. There's a one-to-one correspondence between the two choices, so they have the same count.

What does the factorial symbol (!) mean?

Factorial means multiply all positive integers from 1 up to that number. So 5! = 5 × 4 × 3 × 2 × 1 = 120. By definition, 0! = 1. Factorials grow extremely fast – 10! is already 3,628,800. This is why combination formulas often cancel out large portions of the factorials.

Can r be larger than n?

No, C(n,r) is undefined when r > n. You can't choose 10 items from a set of only 5. Mathematically, the formula would require calculating factorials of negative numbers, which aren't defined. If you try to compute it, you should get 0 or an error.

What is the binomial coefficient?

The binomial coefficient (ⁿ⁄_r) is exactly the same as C(n,r). It's called the binomial coefficient because it appears in the binomial theorem: (a + b)ⁿ = Σ C(n,k) × aⁿ⁻ᵏ × bᵏ. The coefficients in the expansion of (a + b)ⁿ are the combination values.

How do I calculate combinations without a calculator?

Write out the formula and cancel before multiplying. For C(8,3) = 8!/(3!×5!), expand to (8×7×6×5!)/(3!×5!). The 5! cancels, leaving (8×7×6)/(3×2×1). Simplify: 6/6 = 1, so you have 8×7 = 56. Always cancel factorials first to keep numbers manageable.

What's the largest combination I can calculate?

This calculator handles n up to 170 because 171! exceeds the maximum value for standard number types. For most practical purposes, this is more than enough. Lottery calculations, card probabilities, and typical combinatorics problems all fall well within this range.

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