Number of parallelograms when n horizontal parallel lines intersect m vertical parallellines

Given two positive integers n and m. The task is to count number of parallelogram that can be formed of any size when n horizontal parallel lines intersect with m vertical parallel lines.

Examples:

Input : n = 3, m = 2
Output : 3
2 parallelograms of size 1x1 and 1 parallelogram 
of size 2x1.

Input : n = 5, m = 5
Output : 100

The idea is to use Combination, which state, number of ways to choose k items from given n items is given by nCr.
To form a parallelogram, we need two horizontal parallel lines and two vertical parallel lines. So, number of ways to choose two horizontal parallel lines are nC2 and number of ways to choose two vertical parallel lines are mC2. So, total number of possible parallelogram will be nC2 x mC2.

Below is C++ implementation of this approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP Program to find number of parallelogram when 
// n horizontal parallel lines intersect m vertical 
// parallel lines.
#include<bits/stdc++.h>
#define MAX 10
using namespace std;
  
// Find value of Binomial Coefficient
int binomialCoeff(int C[][MAX], int n, int k)
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (int i = 0; i <= n; i++)
    {
        for (int j = 0; j <= min(i, k); j++)
        {
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
   
            // Calculate value using previously
            // stored values
            else
                C[i][j] = C[i-1][j-1] + C[i-1][j];
        }
    }
}
  
// Return number of parallelogram when n horizontal
// parallel lines intersect m vertical parallel lines.
int countParallelogram(int n, int m)
{
    int  C[MAX][MAX] = { 0 };    
    binomialCoeff(C, max(n, m), 2);    
    return C[n][2] * C[m][2];
}
  
// Driver Program
int main()
{
    int n = 5, m = 5;    
    cout << countParallelogram(n, m) << endl;
    return 0;

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to find number of parallelogram when 
// n horizontal parallel lines intersect m vertical 
// parallel lines.
class GFG
{
    static final int MAX = 10;
      
    // Find value of Binomial Coefficient
    static void binomialCoeff(int C[][], int n, int k)
    
        // Calculate value of Binomial Coefficient
        // in bottom up manner
        for (int i = 0; i <= n; i++)
        {
            for (int j = 0; j <= Math.min(i, k); j++)
            {
                // Base Cases
                if (j == 0 || j == i)
                    C[i][j] = 1;
      
                // Calculate value using previously
                // stored values
                else
                    C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
            }
        }
    }
      
    // Return number of parallelogram when n horizontal
    // parallel lines intersect m vertical parallel lines.
    static int countParallelogram(int n, int m)
    {
        int C[][]=new int[MAX][MAX]; 
          
        binomialCoeff(C, Math.max(n, m), 2); 
          
        return C[n][2] * C[m][2];
    }
      
    // Driver code
    public static void main(String arg[]) 
    {
        int n = 5, m = 5
        System.out.println(countParallelogram(n, m));
    }
}
  
// This code is contributed By Anant Agarwal.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python Program to find number of parallelogram when
# n horizontal parallel lines intersect m vertical
# parallel lines.
MAX = 10;
  
# Find value of Binomial Coefficient
def binomialCoeff(C, n, k):
      
    # Calculate value of Binomial Coefficient
    # in bottom up manner
    for i in range(n + 1):
        for j in range(0, min(i, k) + 1):
          
            # Base Cases
            if (j == 0 or j == i):
                C[i][j] = 1;
  
            # Calculate value using previously
            # stored values
            else:
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  
# Return number of parallelogram when n horizontal
# parallel lines intersect m vertical parallel lines.
def countParallelogram(n, m):
    C = [[0 for i in range(MAX)] for j in range(MAX)]
  
    binomialCoeff(C, max(n, m), 2);
  
    return C[n][2] * C[m][2];
  
# Driver code
if __name__ == '__main__':
    n = 5;
    m = 5;
    print(countParallelogram(n, m));
  
# This code is contributed by 29AjayKumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# Program to find number of parallelogram when 
// n horizontal parallel lines intersect m vertical 
// parallel lines.
using System;
  
class GFG
{
    static int MAX = 10;
      
    // Find value of Binomial Coefficient
    static void binomialCoeff(int [,]C, int n, int k)
    
        // Calculate value of Binomial Coefficient
        // in bottom up manner
        for (int i = 0; i <= n; i++)
        {
            for (int j = 0; j <= Math.Min(i, k); j++)
            {
                // Base Cases
                if (j == 0 || j == i)
                    C[i, j] = 1;
      
                // Calculate value using previously
                // stored values
                else
                    C[i, j] = C[i - 1, j - 1] + C[i - 1, j];
            }
        }
    }
      
    // Return number of parallelogram when n horizontal
    // parallel lines intersect m vertical parallel lines.
    static int countParallelogram(int n, int m)
    {
        int [,]C = new int[MAX, MAX]; 
          
        binomialCoeff(C, Math.Max(n, m), 2); 
          
        return C[n, 2] * C[m, 2];
    }
      
    // Driver code
    public static void Main() 
    {
        int n = 5, m = 5; 
        Console.WriteLine(countParallelogram(n, m));
    }
}
  
// This code is contributed By vt_m.

chevron_right


Output:

100

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : 29AjayKumar

Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.