Permutation without Repetition Calculator

Count the ordered arrangements of r items chosen from n distinct items, with no item reused: P(n, r) = n! / (n − r)!.

Quick Facts

Formula
P(n, r) = n! / (n − r)!
Order matters and each item is used at most once. Requires 0 ≤ r ≤ n.

Your Results

Calculated
Permutations P(n, r)
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Ordered arrangements, no repetition
Expanded product
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n × (n−1) × … × (n−r+1)
Combinations C(n, r)
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P(n, r) ÷ r! (order ignored)

Ready

Enter n and r, then press Calculate.

Understanding permutations without repetition

A permutation without repetition counts the number of ways to arrange r items chosen from a set of n distinct items when order matters and no item may be used more than once. It answers questions like "how many different 3-letter codes can I make from 10 distinct letters if no letter repeats?" The answer is denoted P(n, r), nPr, or nPr.

The formula

The number of permutations of r items taken from n distinct items without repetition is:

P(n, r) = n! / (n − r)!

Here n! (n factorial) means n × (n − 1) × (n − 2) × … × 2 × 1, and 0! is defined to equal 1. Dividing n! by (n − r)! cancels the tail of the product, so the formula is equivalent to a short descending product of exactly r factors:

P(n, r) = n × (n − 1) × (n − 2) × … × (n − r + 1)

For example, P(10, 3) = 10 × 9 × 8 = 720. You could also write it as 10! / 7! = 3,628,800 / 5,040 = 720 — the same result, but the descending product is far easier to evaluate by hand and avoids huge intermediate factorials.

Why the formula works

Think of filling r ordered slots one at a time. The first slot can be any of the n items. Because you cannot reuse an item, the second slot has only n − 1 choices left, the third has n − 2, and so on down to the r-th slot, which has n − r + 1 choices. By the multiplication principle you multiply these together, giving exactly r descending factors — which is what n! / (n − r)! computes.

Permutations vs. combinations

Permutations count ordered arrangements, so the sequence ABC is counted separately from CBA. Combinations count unordered selections, where ABC and CBA are the same choice. The two are linked by:

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

Because any set of r items can itself be ordered in r! ways, there are always r! times as many permutations as combinations. If order is irrelevant to your problem (e.g., picking a committee), use combinations; if order matters (e.g., assigning 1st, 2nd, 3rd place), use permutations.

Common reference values

  • P(n, 0) = 1 — there is exactly one way to arrange nothing (the empty arrangement).
  • P(n, 1) = n — choosing and placing a single item has n possibilities.
  • P(n, n) = n! — arranging all items in order is a full permutation; e.g., P(5, 5) = 120.
  • P(52, 5) = 311,875,200 — ordered 5-card deals from a standard deck.
  • P(10, 3) = 720 and P(6, 2) = 30 — handy small cases to memorize.

Frequently Asked Questions

What is the formula for permutations without repetition?
P(n, r) = n! / (n − r)!, where n is the total number of distinct items and r is how many you choose and place in order. Since no item is reused, this equals the descending product n × (n − 1) × … × (n − r + 1) of exactly r factors.
How is a permutation different from a combination?
A permutation counts ordered arrangements (ABC and CBA are different); a combination counts unordered selections (ABC and CBA are the same). They are related by C(n, r) = P(n, r) / r!, so there are always r! more permutations than combinations for the same n and r.
What happens if r is greater than n?
P(n, r) is 0 (undefined in the usual sense) when r > n, because you cannot pick more distinct items than you have without repeating one. This calculator requires 0 ≤ r ≤ n; if you need to allow reuse, that is a permutation with repetition, which uses nr instead.
Why is 0! equal to 1?
Defining 0! = 1 keeps the formulas consistent: it makes P(n, n) = n! / 0! = n! and P(n, 0) = n! / n! = 1, both of which match the combinatorial reality that there is exactly one way to arrange all items and exactly one way to arrange none.

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