Permutations & Combinations Calculator (nPr, nCr)

A combination calculator counts the ways to choose r items from n when the order does not matter: C(n, r) = n! ÷ (r!(n − r)!). Choosing 3 people from 10 gives (10 × 9 × 8) ÷ (3 × 2 × 1) = 120 combinations. If the order matters, as with first, second and third place, the permutation count is P(10, 3) = 720.

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How to use the permutations and combinations calculator

1. Choose what you are counting: combinations (nCr), permutations (nPr), either one with repetition, circular arrangements, or the arrangements of a word with repeated letters.
2. Enter n (how many items you have) and r (how many you choose), or type the word. Optionally, name your items (for example: red, blue, green) to see them in the list.
3. Read the exact answer, then the comparison table that counts the same n and r in five ways. Open Show steps for the working and List all to see the outcomes themselves. Every answer is computed exactly, even with thousands of digits.

Permutation and combination formulas

Here n! (n factorial) means n × (n − 1) × … × 1, and 0! = 1. The only difference between the first two formulas is the extra r! in the combination formula: P(n, r) = C(n, r) × r!, because each group of r items can be put in order in r! ways.

When to use permutations or combinations

Ask two questions. Does order matter? A committee, a pizza’s toppings and a poker hand are combinations; a race podium, a password and a seating plan are permutations. Can an item be used again? Digits in a code can repeat (n^r); people on a team cannot (nCr or nPr). For a round table, rotations of the same seating count once, so use the circular formula.

Worked examples

• Small check: from 5 items, choose 2. C(5, 2) = 10 groups, P(5, 2) = 20 ordered pairs, twice as many because each pair has 2 orders.
• By hand: C(8, 3) = (8 × 7 × 6) ÷ (3 × 2 × 1) = 336 ÷ 6 = 56, and P(8, 3) = 8 × 7 × 6 = 336.
• Poker hands: 5 cards from 52 gives C(52, 5) = 2,598,960 hands.
• Lottery: 6 numbers from 49 gives C(49, 6) = 13,983,816, so one ticket has a 1 in 13,983,816 chance of matching all six.
• Letters of a word: MATH has 4! = 24 arrangements; BOOK has 4! ÷ 2! = 12 because the two O’s are identical; 2-letter arrangements from MATH are P(4, 2) = 12. MISSISSIPPI has 11 letters (I × 4, S × 4, P × 2, M × 1), so 11! ÷ (4! × 4! × 2!) = 39,916,800 ÷ 1,152 = 34,650.

Pascal’s triangle and combinations

Row n of Pascal’s triangle lists C(n, 0), C(n, 1), …, C(n, n). Row 5 is 1, 5, 10, 10, 5, 1. Each entry is the sum of the two above it, C(n, r) = C(n − 1, r − 1) + C(n − 1, r), and each row adds up to 2ⁿ, the number of subsets of n items. The calculator draws the last rows for n up to 20 and highlights your entry.

Combinations in probability

When every outcome is equally likely, probability = favorable outcomes ÷ total outcomes, and both counts are often combinations. From 6 students, a committee of 2 is chosen at random. There are C(6, 2) = 15 possible committees and only one of them is Ana and Ben, so the chance is 1/15. For repeated yes/no trials, C(n, k) gives the number of ways to place k successes, which is the heart of the binomial formula; see the binomial probability calculator and the probability calculator.

Common mistakes

• Using nPr for a group. A team of 3 is the same team in any order, so use nCr.
• Forgetting identical items. Divide by the factorial of each repeat count, as with BOOK and MISSISSIPPI.
• Counting rotations at a round table. 6 people in a row have 720 orders, but only 120 around a table.
• Choosing more than you have. Without repetition r cannot be larger than n; with repetition it can.
• Multiplying out huge factorials. Cancel first: C(52, 5) needs only 52 × 51 × 50 × 49 × 48 ÷ 120.

Permutations and combinations practice quiz: 10 questions

Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.

Frequently asked questions

What is the permutations and combinations formula?

Permutations: P(n, r) = n! ÷ (n − r)!. Combinations: C(n, r) = n! ÷ (r!(n − r)!). For n = 10 and r = 3 they give 720 and 120.

How do you calculate permutations and combinations on a TI-84?

Type n, press MATH, move right to PRB, choose 2:nPr or 3:nCr, then type r and press ENTER. For 10 nCr 3 the screen shows 120.

How do you solve permutations and combinations without a calculator?

Write the formula, cancel the larger factorial, and multiply only the remaining factors. C(10, 3) = (10 × 9 × 8) ÷ (3 × 2 × 1) = 720 ÷ 6 = 120. Using C(n, r) = C(n, n − r) keeps the work short, so C(10, 7) is also 120.

Is permutations and combinations algebra?

It is part of combinatorics, the math of counting, and is usually taught in Algebra 2, precalculus and statistics courses alongside probability.

How do you find the permutations of letters in a word?

Count the letters n and divide n! by the factorial of each letter’s count. BOOKKEEPER has 10 letters with O × 2, K × 2 and E × 3, so 10! ÷ (2! × 2! × 3!) = 151,200 arrangements. Use the Letters of a word tab to do this for any word.

How are permutations and combinations used in probability?

They count the outcomes. The probability of an event is the number of favorable outcomes divided by the total number, such as 1 ÷ C(49, 6) for a lottery jackpot.

Related calculators

• Factorial calculator: n! for large n
• Probability calculator: and, or, not and conditional probability
• Binomial probability calculator: P(X = k) with C(n, k)
• Expected value calculator
• Set calculator: unions, intersections and subsets