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

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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++




// 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;
}

Java




// 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.

Python3




# 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

C#




// 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.

Javascript




<script>
 
// Javascript Program to find number of parallelogram when
// n horizontal parallel lines intersect m vertical
// parallel lines.
var MAX = 10;
 
// Find value of Binomial Coefficient
function binomialCoeff(C, n, k)
{
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (var i = 0; i <= n; i++)
    {
        for (var 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.
function countParallelogram(n, m)
{
    var C = Array.from(Array(MAX), () => Array(MAX).fill(0));
    binomialCoeff(C, Math.max(n, m), 2);   
    return C[n][2] * C[m][2];
}
 
// Driver Program
var n = 5, m = 5;   
document.write( countParallelogram(n, m));
 
// This code is contributed by rdtank.
</script>

Output

100



Time Complexity: O(n2
Auxiliary Space: O(n2)

Approach: Using Basic Maths

The same Question can be Solved By just using the basic maths
as we know nC2 = n*(n-1)/2 and same for mC2 so just using basic maths we can solve the question in O(1)

Below is the implementation of the above approach:

C++




#include <iostream>
 
class GFG {
public:
    static int findtheParallelogram(int n, int m)
    {
        // as nC2 = (n*(n-1))/2
        int result
            = ((n * (n - 1)) / 2) * ((m * (m - 1)) / 2);
 
        return result;
    }
};
 
int main()
{
    int n = 5;
    int m = 5;
    std::cout << GFG::findtheParallelogram(n, m)
              << std::endl;
    return 0;
}

Java




import java.io.*;
 
class GFG {
   public static int findtheParallelogram(int n, int m) {
     //as nC2 = (n*(n-1))/2
    int result = ((n * (n - 1)) / 2) * ((m * (m - 1)) / 2);
      
    return result;
  }
  //Driver code
  public static void main(String[] vars){
    int n = 5;
    int  m =5;
    System.out.println(findtheParallelogram(n,m));
  }
}

Output

100



Time Complexity :O(1)
Space Complexity : O(1)
This article is contributed by Aarti_Rathi and Dhruv Khoradiya. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 


Last Updated : 18 Sep, 2023
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