# Count of rectangles possible from N and M straight lines parallel to X and Y axis respectively

• Last Updated : 06 Sep, 2022

Given two integers N and M, where N straight lines are parallel to the X-axis and M straight lines are parallel to Y-axis, the task is to calculate the number of rectangles that can be formed by these lines.

Examples:

Input: N = 3, M = 6
Output: 45
Explanation:
There are total 45 rectangles possible with 3 lines parallel to x axis and 6 lines parallel to y axis.
Input: N = 2, M = 4
Output:
Explanation:
There are total 6 rectangles possible with 2 lines parallel to x axis and 4 lines parallel to y axis.

Approach:
To solve the problem mentioned above we need to observe that a rectangle is formed by 4 straight lines in which opposite sides are parallel and the angle between any two sides is 90. Hence, for every rectangle, two sides need to be parallel to X-axis and the other two sides need to be parallel to Y-axis.

• Number of ways to select two lines parallel to X axis = NC2 and the Number of ways to select two lines parallel to Y axis = MC2 .
• So the total number of rectangles  = NC2 * MC = [ N * (N – 1) / 2 ] * [ M * (M – 1) / 2 ]

Below is implementation of above approach:

## C++

 `// C++ Program to count number of``// rectangles formed by N lines``// parallel to X axis M lines``// parallel to Y axis``#include ``using` `namespace` `std;` `// Function to calculate``// number of rectangles``int` `count_rectangles(``int` `N, ``int` `M)``{``    ``// Total number of ways to``    ``// select two lines``    ``// parallel to X axis``    ``int` `p_x = (N * (N - 1)) / 2;` `    ``// Total number of ways``    ``// to select two lines``    ``// parallel to Y axis``    ``int` `p_y = (M * (M - 1)) / 2;` `    ``// Total number of rectangles``    ``return` `p_x * p_y;``}` `// Driver Program``int` `main()``{` `    ``int` `N = 3;` `    ``int` `M = 6;` `    ``cout << count_rectangles(N, M);``}`

## Java

 `// Java Program to count number of``// rectangles formed by N lines``// parallel to X axis M lines``// parallel to Y axis``class` `GFG{` `// Function to calculate``// number of rectangles``static` `int` `count_rectangles(``int` `N, ``int` `M)``{``    ``// Total number of ways to``    ``// select two lines``    ``// parallel to X axis``    ``int` `p_x = (N * (N - ``1``)) / ``2``;` `    ``// Total number of ways``    ``// to select two lines``    ``// parallel to Y axis``    ``int` `p_y = (M * (M - ``1``)) / ``2``;` `    ``// Total number of rectangles``    ``return` `p_x * p_y;``}` `// Driver Program``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``3``;``    ``int` `M = ``6``;` `    ``System.out.print(count_rectangles(N, M));``}``}` `// This code is contributed by sapnasingh4991`

## Python3

 `# Python3 program to count number of rectangles``# formed by N lines parallel to X axis``# and M lines parallel to Y axis``def` `count_rectangles(N, M):` `    ``# Total number of ways to select``    ``# two lines parallel to X axis``    ``p_x ``=` `(N ``*` `(N ``-` `1``)) ``/``/` `2` `    ``# Total number of ways to select``    ``# two lines parallel to Y axis``    ``p_y ``=` `(M ``*` `(M ``-` `1``)) ``/``/` `2` `    ``# Total number of rectangles``    ``return` `p_x ``*` `p_y` `# Driver code``N ``=` `3``M ``=` `6` `print``(count_rectangles(N, M))` `# This code is contributed by himanshu77`

## C#

 `// C# Program to count number of``// rectangles formed by N lines``// parallel to X axis M lines``// parallel to Y axis``using` `System;``class` `GFG{` `// Function to calculate``// number of rectangles``static` `int` `count_rectangles(``int` `N, ``int` `M)``{``    ``// Total number of ways to``    ``// select two lines``    ``// parallel to X axis``    ``int` `p_x = (N * (N - 1)) / 2;` `    ``// Total number of ways``    ``// to select two lines``    ``// parallel to Y axis``    ``int` `p_y = (M * (M - 1)) / 2;` `    ``// Total number of rectangles``    ``return` `p_x * p_y;``}` `// Driver Program``public` `static` `void` `Main()``{``    ``int` `N = 3;``    ``int` `M = 6;` `    ``Console.Write(count_rectangles(N, M));``}``}` `// This code is contributed by Code_mech`

## Javascript

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Output:

`45`

Time Complexity: O(1), as constant operations are being performed.
Auxiliary Space: O(1), as constant space is being used.

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