Permutation and Combination Calculator (nPr, nCr)

Enter the total number of items n and the number you choose r. The calculator finds the permutations nPr (order matters) and the combinations nCr (order does not matter) at the same time. The formula below is linked to the input fields, so you can also edit n and r directly in it.

Total items n

Items chosen r

Enter n and r as whole numbers of 0 or more (r must not be larger than n). n can be up to 10000, and every digit of the answer is shown exactly, even when it runs to dozens of digits.

Result

Enter the total number of items n and the number you choose r in the fields on the left and press "Calculate". The result will appear here.

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What you can do on this page
What is this calculation used for?
How to Use
Formula
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

Just enter the total number of items \(n\) and the number you choose \(r\). You get both the permutations \({}_{n}\mathrm{P}_{r}\) (ways to choose when order matters) and the combinations \({}_{n}\mathrm{C}_{r}\) (ways to choose when order does not matter) at once
Answer counting questions such as "How many ways can an 11-player soccer team pick a captain and a goalkeeper?" or "How many ways can it pick 2 forwards?" in one step
Even for a large \(n\) where the answer has dozens or hundreds of digits, every digit is shown exactly, with no rounding
A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page

This page covers permutations and combinations without repetition, where an item cannot be chosen again once it is picked. When the same item can be chosen any number of times, like the faces of a die (permutations or combinations with repetition), the formulas are different, so this calculator does not handle them.

What is this calculation used for?

Putting a number on lottery odds

In Powerball, you pick 5 white balls from 69 and 1 red Powerball from 26. The order of the white balls does not matter, so they are a combination: \({}_{69}\mathrm{C}_{5} = 11{,}238{,}513\) ways. Multiply by the 26 choices for the red ball and you get \(292{,}201{,}338\) possible tickets. Only 1 of them wins the jackpot, so the odds are about 1 in 292 million.
The combination formula lets you check how hard it is to win a lottery or a prize draw with real numbers instead of a vague feeling.

Counting the ways to pick class officers or helpers

In a class of 30 students, choosing a class president and a vice president means the roles are different, so it is a permutation: \({}_{30}\mathrm{P}_{2} = 30 \times 29 = 870\) ways. Just picking 2 students for a cleanup duty, on the other hand, does not depend on order, so it is a combination: \({}_{30}\mathrm{C}_{2} = 435\) ways.
Once you know that the count changes depending on whether roles or order matter, you can count correctly for drawings and elections.

Finding the number of games in a round robin

In a round robin, every team plays every other team once. Each game is a choice of 2 teams, so \(n\) teams play \({}_{n}\mathrm{C}_{2}\) games. With 10 teams, that is \({}_{10}\mathrm{C}_{2} = 45\) games.
Tournament organizers use this calculation to estimate how many games, days and fields they need.

Counting PIN codes and lock codes (security)

If you make a 4-digit code from the 10 digits 0 to 9 without using any digit twice, order matters, so there are \({}_{10}\mathrm{P}_{4} = 5{,}040\) codes. The more codes there are, the harder it is to break the lock by trying them all, and permutations also show how much adding one more digit helps.
By the way, a "combination lock" only opens when the numbers are entered in the right order, so mathematically it is a permutation lock, not a combination lock.

Sampling inspection for quality control (manufacturing)

When 5 products are picked for inspection from a batch of 100, there are \({}_{100}\mathrm{C}_{5} = 75{,}287{,}520\) ways to pick them.
The theory of statistical sampling inspection is built on counts like this. It finds the chance that an inspection catches a defective product in the batch, and it is a key part of quality assurance in factories.

Formula

Factorial \(n!\) (ways to arrange all \(n\) items)

Standard notation (the usual math form)

\(n!\) \(=\) \(n \times (n-1) \times \cdots \times 2 \times 1\)

In words (symbols replaced with words)

② \(n!\): ways to arrange all \(n\) items \(=\) ① the product of the whole numbers from \(n\) down to \(1\)

The formula in words

① Multiply all the whole numbers from \(n\) down to \(1\) together

② and you get the \(n!\): ways to arrange all \(n\) items

Quick example

The number of ways to line up 4 people (A, B, C and D) in a row is

ways to line up all 4 people \(4!\) \(=\) the product from 4 down to 1

\(4! = 4 \times 3 \times 2 \times 1 = 24\)

Key idea

There are \(n\) choices for the first place, then \((n-1)\) choices for the next place from those left, and so on. The number of choices goes down by 1 each time, so multiplying them all gives the number of ways to arrange everything. By definition, \(0! = 1\) (there is exactly 1 way to arrange nothing). Thanks to this rule, the permutation and combination formulas below also work as they are when \(r = n\) or \(r = 0\).

Permutations \({}_{n}\mathrm{P}_{r}\) (order matters)

Standard notation (the usual math form)

\({}_{n}P_{r}\) \(=\) \(n!\) \(\div\) \((n-r)!\)

In words (symbols replaced with words)

③ \({}_{n}\mathrm{P}_{r}\): number of ordered arrangements \(=\) ① \(n!\): ways to arrange all \(n\) items \(\div\) ② \((n-r)!\): ways to arrange the rest

The formula in words

① Take the \(n!\): ways to arrange all \(n\) items

② divide it by the \((n-r)!\): ways to arrange the \((n-r)\) items not chosen

③ and you get the \({}_{n}\mathrm{P}_{r}\): number of ways to choose \(r\) of \(n\) items in order

Quick example

The number of ways to choose a captain and a goalkeeper from the 11 players of a soccer team (the roles are different, so order matters) is

2 of 11 people in order \({}_{11}\mathrm{P}_{2}\) \(=\) ways to arrange all 11 (11!) \(\div\) ways to arrange the other 9 (9!)

\({}_{11}P_{2} = \dfrac{11!}{(11-2)!} = \dfrac{11!}{9!} = 11 \times 10 = 110\)

Key idea

Out of the \(n!\) ways to arrange all \(n\) items, the order of the \((n-r)\) items that were not chosen does not matter, so we divide by \((n-r)!\). That is what this formula means. When you work it out by hand, the fraction cancels down to "multiply \(r\) numbers, counting down from \(n\)". For \({}_{11}\mathrm{P}_{2}\) that is \(11 \times 10 = 110\). A permutation counts "captain A, goalkeeper B" and "captain B, goalkeeper A" as two different choices.

Combinations \({}_{n}\mathrm{C}_{r}\) (order does not matter)

Standard notation (the usual math form)

\({}_{n}C_{r}\) \(=\) \({}_{n}P_{r}\) \(\div\) \(r!\)

In words (symbols replaced with words)

③ \({}_{n}\mathrm{C}_{r}\): number of ways to choose \(=\) ① \({}_{n}\mathrm{P}_{r}\): choices in order \(\div\) ② \(r!\): ways to reorder the \(r\) chosen items

The formula in words

① Take the \({}_{n}\mathrm{P}_{r}\): choices in order

② divide it by the \(r!\): ways to reorder the \(r\) chosen items to remove the repeats caused by order

③ and you get the \({}_{n}\mathrm{C}_{r}\): number of ways to choose \(r\) of \(n\) items

Quick example

The number of ways to choose 2 forwards from the 11 players of a soccer team (the order of the 2 does not matter) is

2 of 11 people \({}_{11}\mathrm{C}_{2}\) \(=\) choices in order (110 ways) \(\div\) ways to reorder 2 people (2! = 2 ways)

\({}_{11}C_{2} = \dfrac{{}_{11}P_{2}}{2!} = \dfrac{110}{2} = 55\)

Key idea

Among the 110 permutations, each pair of people is counted \(2! = 2\) times, once as "A and B" and once as "B and A". Dividing by \(r!\) removes these repeats and leaves only the number of different groups. Substitute the permutation formula \({}_{n}\mathrm{P}_{r} = \dfrac{n!}{(n-r)!}\) and you get the familiar textbook form \({}_{n}\mathrm{C}_{r} = \dfrac{n!}{r!\,(n-r)!}\). \({}_{n}\mathrm{C}_{r}\) is also called a binomial coefficient and is often written \(\binom{n}{r}\), read "n choose r".

The permutations nPr count the ways to choose when order matters, and the combinations nCr count the ways to choose when order does not matter. The easiest way to remember them is this link - find the permutations first, then divide by \(r!\) to remove the repeats caused by order, and you have the combinations.

Symbols and terms

Symbols

\(n\)	en	The total number of items you choose from. (Example - choosing from 11 people, \(n = 11\))
\(r\)	ar	The number of items you choose from the total. (Example - choosing 2 people, \(r = 2\))
\(n!\)	n factorial	The product of all the whole numbers from \(n\) down to \(1\). It is the number of ways to arrange all \(n\) items in a row. By definition, \(0! = 1\). (Example - \(4! = 4 \times 3 \times 2 \times 1 = 24\))
\({}_{n}\mathrm{P}_{r}\)	n P r	The number of ways to choose \(r\) of \(n\) items and put them in order (permutations). P stands for "permutation". It is also written \(P(n, r)\).
\({}_{n}\mathrm{C}_{r}\)	n C r	The number of ways to choose \(r\) of \(n\) items (combinations). C stands for "combination". It is also written \(C(n, r)\).
\(\binom{n}{r}\)	n choose r (binomial coefficient)	Another way to write \({}_{n}\mathrm{C}_{r}\). This is the form most often used in college math and in many textbooks.

Terms

counting	Finding how many possible outcomes there are in total. Permutations and combinations are both tools for counting.
permutation	A way to choose \(r\) of \(n\) items and put them in a row, where order matters. Different roles or ranks, such as "captain and vice captain", are counted as different.
combination	A way to choose \(r\) of \(n\) items where order does not matter. As with "2 people for a cleanup duty", the same group of people counts as 1 way.
factorial	Multiplying all the whole numbers from \(n\) down to \(1\). The symbol is \(n!\), and it gives the number of ways to arrange all \(n\) items.
binomial coefficient	Another name for \({}_{n}\mathrm{C}_{r}\). The name comes from the fact that the coefficients you get when you expand \((a+b)^n\) are exactly the values of \({}_{n}\mathrm{C}_{r}\).
tree diagram	A branching diagram that lists every possible outcome, so you count each one exactly once. The permutation and combination formulas do the same counting as a tree diagram in a single calculation.
permutation with repetition	A permutation where the same item can be chosen any number of times. There are \(n^r\) of them. The calculator on this page only handles the case without repetition.
combination with repetition	A combination where the same item can be chosen any number of times. The calculator on this page only handles the case without repetition.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.

Multiplication and division (Grades 3–5)	Being able to multiply whole numbers, as in \(11 \times 10\), and divide, as in \(110 \div 2\)
Counting outcomes and tree diagrams (Grade 7)	Being able to list all possible outcomes with a tree diagram or similar, counting each one exactly once Seeing that arranging (order matters) and choosing (order does not matter) are two different ways of counting
Factorial notation (high school, Algebra 2 or Precalculus)	Knowing that \(n!\) means "multiply everything from \(n\) down to \(1\)" Knowing the rule that \(0! = 1\)
Permutations and combinations (high school, Algebra 2 or Statistics)	Choosing correctly between them - the permutations nPr when order matters, the combinations nCr when it does not Being able to explain that combinations are "permutations divided by \(r!\) to remove the repeats caused by order"

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table to find the factorial n!

Items to arrange n	4
All arrangements n!	=FACT(B1)

Table to find the permutations nPr

Total items n	11
Items chosen r	2
Permutations nPr	=PERMUT(B1,B2)

Table to find the combinations nCr

Total items n	11
Items chosen r	2
Combinations nCr	=COMBIN(B1,B2)

After pasting, B1 (and B2) are your inputs and the last row is calculated automatically.
FACT, PERMUT and COMBIN are the Excel functions for the factorial, permutations and combinations.
For example, B3 shows 110 in the second table and 55 in the third. Just replace B1 and B2 with your own numbers.
Note that Excel keeps only 15 significant digits, so when n is large and the answer has dozens of digits, the last digits are rounded.

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table to find the factorial n!

Items to arrange n	4
All arrangements n!	=FACT(B1)

Table to find the permutations nPr

Total items n	11
Items chosen r	2
Permutations nPr	=PERMUT(B1,B2)

Table to find the combinations nCr

Total items n	11
Items chosen r	2
Combinations nCr	=COMBIN(B1,B2)

The same formulas as in Excel work as is (FACT, PERMUT and COMBIN have the same names in Google Sheets).
Copy the whole table, paste it into cell A1, and replace B1 and B2 with your own numbers.

How to calculate it in Python

import math

total_items = 11    # total number of items n
chosen_items = 2    # number of items chosen r

permutations = math.perm(total_items, chosen_items)   # permutations nPr
combinations = math.comb(total_items, chosen_items)   # combinations nCr

print(f"Permutations nPr (order matters): {permutations}")
print(f"Combinations nCr (order does not matter): {combinations}")

Runs with just the standard math library (math.perm and math.comb need Python 3.8 or later). Python integers have no limit on the number of digits, so the answer stays exact even with hundreds of digits. Change the first two numbers and run it.

How to write it in LaTeX and other math languages (copy and paste)

Factorial \(n!\) (ways to arrange all \(n\) items)

n! = n × (n − 1) × ⋯ × 2 × 1

n! = n \times (n-1) \times \cdots \times 2 \times 1

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>n</mi><mo>!</mo>
    <mo>=</mo>
    <mi>n</mi>
    <mo>&#xD7;</mo>
    <mo>(</mo><mi>n</mi><mo>&#x2212;</mo><mn>1</mn><mo>)</mo>
    <mo>&#xD7;</mo>
    <mo>&#x22EF;</mo>
    <mo>&#xD7;</mo>
    <mn>2</mn>
    <mo>&#xD7;</mo>
    <mn>1</mn>
  </mrow>
</math>

n! = n xx (n-1) xx cdots xx 2 xx 1

Factorial[n]

nFactorial := factorial(n);

n_factorial = factorial(n);

n! = n × (n − 1) × ⋯ × 2 × 1

Permutations \({}_{n}\mathrm{P}_{r}\) (order matters)

ₙPᵣ = n! ÷ (n − r)!

{}_{n}P_{r} = \dfrac{n!}{(n-r)!}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mmultiscripts>
      <mi>P</mi>
      <mi>r</mi><none/>
      <mprescripts/>
      <mi>n</mi><none/>
    </mmultiscripts>
    <mo>=</mo>
    <mfrac>
      <mrow><mi>n</mi><mo>!</mo></mrow>
      <mrow><mo>(</mo><mi>n</mi><mo>&#x2212;</mo><mi>r</mi><mo>)</mo><mo>!</mo></mrow>
    </mfrac>
  </mrow>
</math>

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

n!/(n - r)!

nPr := factorial(n)/factorial(n - r);

npr = factorial(n)/factorial(n - r);

P(n,r) = n!/(n − r)!

Combinations \({}_{n}\mathrm{C}_{r}\) (order does not matter)

ₙCᵣ = n! ÷ (r! × (n − r)!)

{}_{n}C_{r} = \dfrac{n!}{r!\,(n-r)!}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mmultiscripts>
      <mi>C</mi>
      <mi>r</mi><none/>
      <mprescripts/>
      <mi>n</mi><none/>
    </mmultiscripts>
    <mo>=</mo>
    <mfrac>
      <mrow><mi>n</mi><mo>!</mo></mrow>
      <mrow>
        <mi>r</mi><mo>!</mo>
        <mo>(</mo><mi>n</mi><mo>&#x2212;</mo><mi>r</mi><mo>)</mo><mo>!</mo>
      </mrow>
    </mfrac>
  </mrow>
</math>

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

Binomial[n, r]

nCr := binomial(n, r);

ncr = nchoosek(n, r);

C(n,r) = n!/(r!(n − r)!)

How to have ChatGPT do the calculation

You are a calculation assistant for counting (permutations and combinations). Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

A soccer team has 11 players.
1. How many ways are there to choose a captain and a goalkeeper? (The roles are different, so this is a permutation.)
2. How many ways are there to choose 2 forwards? (The order of the 2 does not matter, so this is a combination.)

For each question, say whether you used the permutations nPr or the combinations nCr, and show the numbers from the execution result.

How to Use

1

Enter your numbers

Type the numbers you want to calculate with into the input fields
2

Calculate

Press the "Calculate" button
3

Check the result

The result appears on the spot. The same page also explains the idea behind the calculation and the formula