Given two numbers n and r, the task is to find the value of nPr.

**nPr** represents n permutation r which is calculated as **n!/(n-k)!**. Permutation refers to the process of arranging all the members of a given set to form a sequence. The number of permutations on a set of n elements is given by n!, where “!” represents factorial.

^{n}P_{r}= n! / (n - r)!

**Program:**

## C++

`// CPP program to calculate nPr ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `fact(` `int` `n) ` `{ ` ` ` `if` `(n <= 1) ` ` ` `return` `1; ` ` ` `return` `n * fact(n - 1); ` `} ` ` ` `int` `nPr(` `int` `n, ` `int` `r) ` `{ ` ` ` `return` `fact(n) / fact(n - r); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 5; ` ` ` `int` `r = 2; ` ` ` ` ` `cout << n << ` `"P"` `<< r << ` `" = "` `<< nPr(n, r); ` `} ` ` ` `// This code is contributed by ` `// Surendra_Gangwar ` |

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## Java

`// Java program to calculate nPr ` ` ` `import` `java.util.*; ` ` ` `public` `class` `GFG { ` ` ` ` ` `static` `int` `fact(` `int` `n) ` ` ` `{ ` ` ` `if` `(n <= ` `1` `) ` ` ` `return` `1` `; ` ` ` `return` `n * fact(n - ` `1` `); ` ` ` `} ` ` ` ` ` `static` `int` `nPr(` `int` `n, ` `int` `r) ` ` ` `{ ` ` ` `return` `fact(n) / fact(n - r); ` ` ` `} ` ` ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `int` `n = ` `5` `; ` ` ` `int` `r = ` `2` `; ` ` ` ` ` `System.out.println(n + ` `"P"` `+ r + ` `" = "` ` ` `+ nPr(n, r)); ` ` ` `} ` `} ` |

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## Python3

`# Python3 program to calculate nPr ` `import` `math ` `def` `fact(n): ` ` ` `if` `(n <` `=` `1` `): ` ` ` `return` `1` ` ` ` ` `return` `n ` `*` `fact(n ` `-` `1` `) ` ` ` `def` `nPr(n, r): ` ` ` ` ` `return` `math.floor(fact(n) ` `/` ` ` `fact(n ` `-` `r)) ` ` ` `# Driver code ` `n ` `=` `5` `r ` `=` `2` ` ` `print` `(n, ` `"P"` `, r, ` `"="` `, nPr(n, r)) ` ` ` `# This code contributed by Rajput-Ji ` |

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## C#

`// C# program to calculate nPr ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `static` `int` `fact(` `int` `n) ` ` ` `{ ` ` ` `if` `(n <= 1) ` ` ` `return` `1; ` ` ` `return` `n * fact(n - 1); ` ` ` `} ` ` ` ` ` `static` `int` `nPr(` `int` `n, ` `int` `r) ` ` ` `{ ` ` ` `return` `fact(n) / fact(n - r); ` ` ` `} ` ` ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `n = 5; ` ` ` `int` `r = 2; ` ` ` ` ` `Console.WriteLine(n + ` `"P"` `+ r + ` `" = "` ` ` `+ nPr(n, r)); ` ` ` `} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */` |

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## PHP

`<?php ` `// PHP program to calculate nPr ` `function` `fact(` `$n` `) ` `{ ` ` ` `if` `(` `$n` `<= 1) ` ` ` `return` `1; ` ` ` ` ` `return` `$n` `* fact(` `$n` `- 1); ` `} ` ` ` `function` `nPr(` `$n` `, ` `$r` `) ` `{ ` ` ` `return` `floor` `(fact(` `$n` `) / ` ` ` `fact(` `$n` `- ` `$r` `)); ` `} ` ` ` `// Driver code ` `$n` `= 5; ` `$r` `= 2; ` ` ` `echo` `$n` `, ` `"P"` `, ` `$r` `, ` `" = "` `, nPr(` `$n` `, ` `$r` `); ` ` ` `// This code is contributed by Ryuga ` `?> ` |

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**Output:**

5P2 = 20

Optimization for multiple queries of nPr

If there are multiple queries for nPr, we may precompute factorial values and use same for every call. This would avoid computation of same factorial values again and again.

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