# Number of rectangles in N*M grid

We are given a N*M grid, print the number of rectangles in it.

Examples:

Input : N = 2, M = 2 Output : 9 There are 4 rectangles of size 1 x 1. There are 2 rectangles of size 1 x 2 There are 2 rectangles of size 2 x 1 There is one rectangle of size 1 x 1. Input : N = 5, M = 4 Output : 150 Input : N = 4, M = 3 Output: 60

We have discussed counting number of squares in a n x m grid,

Let us derive a formula for number of rectangles.

If the grid is 1×1, there is 1 rectangle.

If the grid is 2×1, there will be 2 + 1 = 3 rectangles

If it grid is 3×1, there will be 3 + 2 + 1 = 6 rectangles.

we can say that for N*1 there will be N + (N-1) + (n-2) … + 1 = (N)(N+1)/2 rectangles

If we add one more column to N×1, firstly we will have as many rectangles in the 2nd column as the first,

and then we have that same number of 2×M rectangles.

So N×2 = 3 (N)(N+1)/2

After deducing this we can say

For N*M we’ll have (M)(M+1)/2 (N)(N+1)/2 =** M(M+1)(N)(N+1)/4**

So the formula for total rectangles will be M(M+1)(N)(N+1)/4

## C++

`// C++ program to count number of rectangles ` `// in a n x m grid ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `rectCount(` `int` `n, ` `int` `m) ` `{ ` ` ` `return` `(m * n * (n + 1) * (m + 1)) / 4; ` `} ` ` ` `/* driver code */` `int` `main() ` `{ ` ` ` `int` `n = 5, m = 4; ` ` ` `cout << rectCount(n, m); ` ` ` `return` `0; ` `} ` |

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## Java

`// JAVA Code to count number of ` `// rectangles in N*M grid ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ` `public` `static` `long` `rectCount(` `int` `n, ` `int` `m) ` ` ` `{ ` ` ` `return` `(m * n * (n + ` `1` `) * (m + ` `1` `)) / ` `4` `; ` ` ` `} ` ` ` ` ` `/* Driver program to test above function */` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `n = ` `5` `, m = ` `4` `; ` ` ` `System.out.println(rectCount(n, m)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. ` |

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## Python3

`# Python3 program to count number ` `# of rectangles in a n x m grid ` ` ` `def` `rectCount(n, m): ` ` ` ` ` `return` `(m ` `*` `n ` `*` `(n ` `+` `1` `) ` `*` `(m ` `+` `1` `)) ` `/` `/` `4` ` ` `# Driver code ` `n, m ` `=` `5` `, ` `4` `print` `(rectCount(n, m)) ` ` ` `# This code is contributed by Anant Agarwal. ` |

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## C#

`// C# Code to count number of ` `// rectangles in N*M grid ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `public` `static` `long` `rectCount(` `int` `n, ` `int` `m) ` ` ` `{ ` ` ` `return` `(m * n * (n + 1) * (m + 1)) / 4; ` ` ` `} ` ` ` ` ` `// Driver program ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `n = 5, m = 4; ` ` ` `Console.WriteLine(rectCount(n, m)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Anant Agarwal. ` |

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## PHP

`<?php ` `// PHP program to count ` `// number of rectangles ` `// in a n x m grid ` ` ` `function` `rectCount(` `$n` `, ` `$m` `) ` `{ ` ` ` `return` `(` `$m` `* ` `$n` `* ` ` ` `(` `$n` `+ 1) * ` ` ` `(` `$m` `+ 1)) / 4; ` `} ` ` ` `// Driver Code ` `$n` `= 5; ` `$m` `= 4; ` `echo` `rectCount(` `$n` `, ` `$m` `); ` ` ` `// This code is contributed ` `// by ajit ` `?> ` |

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**Output:**

150

This article is contributed by **Pranav**. 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.

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