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Permutation without Repetition Calculator

Created by Aleksandra Zając, MD
Reviewed by Steven Wooding
Last updated: Jan 18, 2024

If you want to know how many different ways to choose r elements from the set of n elements, this permutation without repetition calculator is a good place for you! Reading the text, you'll learn:

  • Permutation without repetition formula;
  • How many permutations of 5 numbers can you get; and
  • How many different ways (permutations – nPr) are there to choose any desired number of elements from any desired set?

What is permutation without repetition?

A permutation is a specific selection of elements within a set where the order of the elements is essential. We call a similar selection but without regard for the order a combination.

We can differentiate two different types of permutations:

  • With repetition – when you can choose again the element that was previously chosen; and
  • Without repetition – when you can't.

Let's take a simple example to illustrate this.

We've got a set of two elements – A and B. We want to derive 2 elements each time. How many permutations and combinations can we get?

  • Combinations (order doesn't matter) – just one, AB.
  • Combinations with repetition – three, AB, AA, BB.
  • Permutations without repetition – two, AB and BA.
  • Permutations with repetition – four: AB, BA, AA, BB.

How to calculate a number of permutations without repetition (nPR)?

While playing with small numbers, the number of permutations is easy to figure out. But you cannot just sit with paper and pencil for bigger numbers and write all of them down. Luckily, we've got a particular permutation without a repetition formula at hand!

P(n,r)=n!(nr)!\small P(n, r) = \frac{n!}{(n-r)!}


  • P(n,r)P(n, r) – Number of permutations (also called nPr);
  • nn – Number of elements in the set; and
  • rr – Number of elements you choose from the set.

The !! – exclamation mark – is a symbol for. Take a look on the factorial calculator to dig deeper.

Let's put that into practice. Imagine we've got a set of 10 different elements, and we want to choose 5 of them. In how many different ways can we get the set of 5 elements?

  • nn = 10; and
  • rr = 5.
P(n,r)=10!(105)!=10!5!=3628800120=30240\small \begin{split} P(n, r) &= \frac{10!}{(10-5)!}\\[1.2em] &= \frac{10!}{5!} = \frac{3628800}{120}\\[1em] &= 30240 \end{split}

Answer: with a set of 10 elements and choosing five of them, we can get 30240 different permutations without repetition.

How to use permutation without repetition calculator?

Our tool is comfortable and easy to use. Follow these steps:

  1. The calculator panel on your left consists of 3 sections.

  2. In the first section, you enter your data. Fill in the first row with number of objects (n) you've got.

  3. Enter the number of objects you want to choose (r).

  4. Move to the second section. After filling in the first two rows, this part contains your results already. In the first row, you'll see permutations without repetitions – this is the field that interests you.

  5. The third part of our permutation without repetition calculator will present the results visually. You can switch mode from Permutations to the others – combinations, and both permutations and combinations with repetitions.

Our permutation tools

Check out the rest of our permutation tools:

  1. Permutation calculator; and
  2. Permutation with repetition calculator.


How many permutations of 5 numbers can you make?

There are 120 permutations without repetitions that you can derive from a 5-number set if you want to get 5-digits permutations.

How do I calculate the number of permutations (nPr) for a set of 11 elements?

To calculate the number of permutations – nPr:

  1. Determine how many numbers you want to choose from the original set. Assume it's 4.

  2. Use the permutation without repetition formula:

    nPr= n!/(n - r)!

    where: n – number of elements in the set (11); and r – number of elements to choose (4).

  3. Solve the equation: nPr = 11!/(11 - 4!) = 11!/7! = 39916800/5040 = 7920.

  4. The number of permutations equals 7920.

Aleksandra Zając, MD
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