Permutation & Combination Calculator — Step-by-Step Solutions

The Permutation & Combination Calculator computes the number of ways to arrange or select items from a set. It calculates permutations (order matters) and combinations (order does not matter) along with factorials.

Enter n (total items) and r (items chosen) to get instant results with step-by-step working. The calculator handles large numbers and shows the formulas used for each calculation.

n (total items)

r (chosen items)

Your calculations will appear here

How to Use the Permutation & Combination Calculator

Select the calculation mode: Permutation (nPr), Combination (nCr), or Factorial (n!).
Enter n, the total number of items in the set.
Enter r, the number of items being chosen (not needed for factorial mode).
Review the result along with the step-by-step formula application and intermediate calculations.

Formula and Theory

A permutation counts the number of ways to arrange r items from a set of n items where order matters. The formula is P(n, r) = n! / (n - r)!. For example, arranging 3 people out of 10 into first, second, and third place gives P(10, 3) = 10 * 9 * 8 = 720 distinct arrangements.

A combination counts the number of ways to select r items from a set of n items where order does not matter. The formula is C(n, r) = n! / (r! * (n - r)!). Selecting 3 people out of 10 for a committee (where positions are identical) gives C(10, 3) = 120.

The factorial n! is the product of all positive integers from 1 to n. It grows extremely rapidly: 10! = 3,628,800 and 20! exceeds 2.4 * 10^18. By convention, 0! = 1. Factorials are the building blocks of both permutation and combination formulas.

Worked Examples

Example 1: Permutation: arranging podium finishers

Problem: In a race with 8 runners, how many different ways can gold, silver, and bronze be awarded?

Solution: P(8, 3) = 8! / (8-3)! = 8! / 5! = 8 * 7 * 6 = 336.

Answer: 336 arrangements

Example 2: Combination: forming a committee

Problem: How many ways can you choose 4 members from a club of 15?

Solution: C(15, 4) = 15! / (4! * 11!) = (15 * 14 * 13 * 12) / (4 * 3 * 2 * 1) = 1365.

Answer: 1,365 ways

Example 3: Factorial calculation

Problem: Calculate 7!

Solution: 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040.

Answer: 5,040

Example 4: Real-world: lottery odds

Problem: A lottery draws 6 numbers from 49. How many possible combinations exist?

Solution: C(49, 6) = 49! / (6! * 43!) = (49 * 48 * 47 * 46 * 45 * 44) / 720 = 13,983,816.

Answer: 13,983,816 possible combinations

Frequently Asked Questions

What is the difference between a permutation and a combination?

In a permutation, the order of selection matters (e.g., first place vs second place). In a combination, only the selection itself matters, not the order. This is why C(n, r) is always less than or equal to P(n, r).

When should I use permutations?

Use permutations when the arrangement or sequence matters. Examples include ranking contestants, assigning numbered positions, or creating passwords where the order of characters is significant.

When should I use combinations?

Use combinations when you are simply choosing a subset and the order is irrelevant. Examples include selecting lottery numbers, picking team members, or choosing items from a menu.

Why is 0! equal to 1?

By convention, 0! = 1 because there is exactly one way to arrange zero objects: do nothing. This definition also ensures that the formulas for permutations and combinations work correctly when r = 0 or r = n.

Can permutations and combinations handle large numbers?

Yes, though the values grow very quickly. For extremely large n, calculators use logarithms or the Stirling approximation to handle the computation. This calculator supports values up to the limits of JavaScript number precision.