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; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

24

## Recommended Posts:

- Total number of ways to place X and Y at n places such that no two X are together
- Number of ways to place two queens on a N*N chess-board
- Count ways to distribute m items among n people
- Count the number of ways to fill K boxes with N distinct items
- Count of ways to distribute N items among 3 people with one person receiving maximum
- Program to find the profit or loss when CP of N items is equal to SP of M items
- Probability of distributing M items among X bags such that first bag contains N items
- Place Value of a given digit in a number
- Minimum numbers with one's place as 9 to be added to get N
- Number of different positions where a person can stand
- Find a number containing N - 1 set bits at even positions from the right
- Find the unit place digit of sum of N factorials
- Count subarrays having sum of elements at even and odd positions equal
- Find a way to fill matrix with 1's and 0's in blank positions
- Sort all special primes in their relative positions
- Minimum cost to cover the given positions in a N*M grid
- Number of positions such that adding K to the element is greater than sum of all other elements
- Count of Numbers in a Range divisible by m and having digit d in even positions
- Puzzle | Place numbers 1 to 9 in a Circle such that sum of every triplet in straight line is 15
- Count Numbers in Range with difference between Sum of digits at even and odd positions as Prime

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.