Calculate nCr using Pascal’s Triangle

A useful application of Pascal’s triangle is the calculation of combinations. The formula to find nCr is n! / r! * (n – r)! which is also the formula for a cell of Pascal’s triangle.

Input: n = 5, r = 3
Output: 10
n! / r! * (n – r)! = 5! / 3! * (2!) = 120 / 12 = 10

Input: n = 7, r = 2
Output: 21



Approach: The idea is to store the Pascal’s triangle in a matrix then the value of nCr will be the value of the cell at nth row and rth column.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Initialize the matrix with 0
int l[1001][1001] = { 0 };
  
int initialize()
{
  
    // 0C0 = 1
    l[0][0] = 1;
    for (int i = 1; i < 1001; i++) {
        // Set every nCr = 1 where r = 0
        l[i][0] = 1;
        for (int j = 1; j < i + 1; j++) {
  
            // Value for the current cell of Pascal's triangle
            l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]);
        }
    }
}
  
// Function to return the value of nCr
int nCr(int n, int r)
{
    // Return nCr
    return l[n][r];
}
  
// Driver code
int main()
{
    // Build the Pascal's triangle
    initialize();
    int n = 8;
    int r = 3;
    cout << nCr(n, r);
}
  
// This code is contributed by ihritik

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Java














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// Java implementation of the approach 


  


class GFG { 


    // Initialize the matrix with 0 


    static int l[][] = new int[1001][1001]; 


  


    static void initialize() 


    { 


  


        // 0C0 = 1 


        l[0][0] = 1; 


        for (int i = 1; i < 1001; i++) { 


            // Set every nCr = 1 where r = 0 


            l[i][0] = 1; 


            for (int j = 1; j < i + 1; j++) { 


  


                // Value for the current cell of Pascal's triangle 


                l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]); 


            } 


        } 


    } 


  


    // Function to return the value of nCr 


    static int nCr(int n, int r) 


    { 


        // Return nCr 


        return l[n][r]; 


    } 


    // Driver code 


    public static void main(String[] args) 


    { 


        // Build the Pascal's triangle 


        initialize(); 


        int n = 8; 


        int r = 3; 


        System.out.println(nCr(n, r)); 


    } 


} 


  


// This code is contributed by ihritik 



















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Python3














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# Python3 implementation of the approach 


  


# Initialize the matrix with 0 


l = [[0 for i in range(1001)] for j in range(1001)] 


  


def initialize(): 


      


    # 0C0 = 1 


    l[0][0] = 1


    for i in range(1, 1001): 


          


        # Set every nCr = 1 where r = 0 


        l[i][0] = 1


        for j in range(1, i + 1): 


              


            # Value for the current cell of Pascal's triangle 


            l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]) 


  


# Function to return the value of nCr 


def nCr(n, r):  


    # Return nCr 


    return l[n][r] 


  


# Driver code 


# Build the Pascal's triangle 


initialize() 


n = 8


r = 3


print(nCr(n, r)) 



















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C#














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// C# implementation of the approach 


  


using System; 


class GFG { 


    // Initialize the matrix with 0 


    static int[, ] l = new int[1001, 1001]; 


  


    static void initialize() 


    { 


  


        // 0C0 = 1 


        l[0, 0] = 1; 


        for (int i = 1; i < 1001; i++) { 


            // Set every nCr = 1 where r = 0 


            l[i, 0] = 1; 


            for (int j = 1; j < i + 1; j++) { 


  


                // Value for the current cell of Pascal's triangle 


                l[i, j] = (l[i - 1, j - 1] + l[i - 1, j]); 


            } 


        } 


    } 


  


    // Function to return the value of nCr 


    static int nCr(int n, int r) 


    { 


        // Return nCr 


        return l[n, r]; 


    } 


    // Driver code 


    public static void Main() 


    { 


        // Build the Pascal's triangle 


        initialize(); 


        int n = 8; 


        int r = 3; 


        Console.WriteLine(nCr(n, r)); 


    } 


} 


  


// This code is contributed by ihritik 



















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Output:


56






          
          
          

          

    

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