Given an integer **N** where **4 ≤ N ≤ 100**. There are **N** lines vertically and **N** lines horizontally. So, There are **N ^{2}** intersections. The task is find the number of ways to place 4 items in these

**N**positions such that each row and column contain not more than one item.

^{2}**Examples:**

Input:N = 4

Output:24

Input:N = 5

Output:600

**Approach:** The number of ways to choose 4 horizontal lines that will have items from **n** is ** ^{n}C_{4}**. There are

**n**ways to place an item on the first of these lines. Given the place of the first item, there are

**n – 1**ways to place an item on the second of these lines because one of the vertical lines is already taken. Given the places of the first and second items, there are

**n – 2**ways to place an item on the third line and the same way

**n – 3**for the fourth item. The total number of ways to place items on selected 4 horizontal paths is

**n * (n – 1) * (n – 2) * (n – 3)**. So the result is

**.**

^{n}C_{4}* n * (n – 1) * (n – 2) * (n – 3)Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the number of ways ` `// to place 4 items in n^2 positions ` `long` `long` `NumberofWays(` `int` `n) ` `{ ` ` ` `long` `long` `x = (1LL * (n) * (n - 1) * (n - 2) * (n - 3)) ` ` ` `/ (4 * 3 * 2 * 1); ` ` ` `long` `long` `y = (1LL * (n) * (n - 1) * (n - 2) * (n - 3)); ` ` ` ` ` `return` `(1LL * x * y); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 4; ` ` ` `cout << NumberofWays(n); ` ` ` ` ` `return` `0; ` `} ` |

## Java

`// Java implementation of the approach ` `class` `GFG ` `{ ` ` ` `// Function to return the number of ways ` `// to place 4 items in n^2 positions ` `static` `long` `NumberofWays(` `int` `n) ` `{ ` ` ` `long` `x = (1l * (n) * (n - ` `1` `) * (n - ` `2` `) * (n - ` `3` `)) ` ` ` `/ (` `4` `* ` `3` `* ` `2` `* ` `1` `); ` ` ` `long` `y = (1l * (n) * (n - ` `1` `) * (n - ` `2` `) * (n - ` `3` `)); ` ` ` ` ` `return` `(1l * x * y); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` ` ` `int` `n = ` `4` `; ` ` ` `System.out.println( NumberofWays(n)); ` `} ` `} ` ` ` `// This code is contributed by Arnab Kundu ` |

## Python3

`# python implementation of the approach ` ` ` `# Function to return the number of ways ` `# to place 4 items in n^2 positions ` `def` `NumbersofWays(n): ` ` ` `x ` `=` `(n ` `*` `(n ` `-` `1` `) ` `*` `(n ` `-` `2` `) ` `*` `(n ` `-` `3` `)) ` `/` `/` `(` `4` `*` `3` `*` `2` `*` `1` `) ` ` ` `y ` `=` `n ` `*` `(n ` `-` `1` `) ` `*` `(n ` `-` `2` `) ` `*` `(n ` `-` `3` `) ` ` ` ` ` `return` `x ` `*` `y ` ` ` `# Driver code ` `n ` `=` `4` `print` `(NumbersofWays(n)) ` ` ` `# This code is contributed by Shrikant13 ` |

## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to return the number of ways ` `// to place 4 items in n^2 positions ` `public` `static` `long` `NumberofWays(` `int` `n) ` `{ ` ` ` `long` `x = (1l * (n) * (n - 1) * (n - 2) * ` ` ` `(n - 3)) / (4 * 3 * 2 * 1); ` ` ` `long` `y = (1l * (n) * (n - 1) * (n - 2) * ` ` ` `(n - 3)); ` ` ` ` ` `return` `(1l * x * y); ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(` `string` `[] args) ` `{ ` ` ` `int` `n = 4; ` ` ` `Console.WriteLine(NumberofWays(n)); ` `} ` `} ` ` ` `// This code is contributed by Shrikant13 ` |

## PHP

`<?php ` `// PHP implementation of the approach ` ` ` `// Function to return the number of ways ` `// to place 4 items in n^2 positions ` `function` `NumberofWays(` `$n` `) ` `{ ` ` ` `$x` `= (1 * (` `$n` `) * (` `$n` `- 1) * ` ` ` `(` `$n` `- 2) * (` `$n` `- 3)) / ` ` ` `(4 * 3 * 2 * 1); ` ` ` `$y` `= (1 * (` `$n` `) * (` `$n` `- 1) * ` ` ` `(` `$n` `- 2) * (` `$n` `- 3)); ` ` ` ` ` `return` `(1 * ` `$x` `* ` `$y` `); ` `} ` ` ` `// Driver code ` `$n` `= 4; ` `echo` `NumberofWays(` `$n` `); ` ` ` `// This code is contributed by mits ` `?> ` |

**Output:**

24