# Number of rectangles in N*M grid

• Difficulty Level : Hard
• Last Updated : 13 Jun, 2022

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+1C2 * M+1C2 = (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 ``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

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## Javascript

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Output:

`150`

Time complexity: O(1)

Auxiliary Space: O(1)

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