# 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 2 x 2. 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

.

**Combinatorial Logic:**

N*M grid can be represented as (N+1) vertical lines and (M+1) horizontal lines.

In a rectangle, we need two distinct horizontal and two distinct verticals.

So going by the logic of Combinatorial Mathematics we can choose 2 vertical lines and 2 horizontal lines to form a rectangle. And total number of these combinations is the number of rectangles possible in the grid.

Total Number of Rectangles in N*M grid: ** ^{N+1}C_{2} * ^{M+1}C_{2 }**=

**(N*(N+1)/2!)*(M*(M+1)/2!)**=

**N*(N+1)*M*(M+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;` `}` |

## 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.` |

## 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.` |

## 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.` |

## 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` `?>` |

## Javascript

`<script>` ` ` `// Javascript Code to count number` ` ` `// of rectangles in N*M grid` ` ` ` ` `function` `rectCount(n, m)` ` ` `{` ` ` `return` `parseInt((m * n * (n + 1) *` ` ` `(m + 1)) / 4, 10);` ` ` `}` ` ` ` ` `let n = 5, m = 4;` ` ` `document.write(rectCount(n, m));` ` ` `</script>` |

**Output:**

150

**Time complexity:** O(1)

**Auxiliary Space: **O(1), since no extra space has been taken.

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