# 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

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