Program to calculate value of nCr using Recursion

Given two numbers N and r, find the value of ^NC_r using recursion

Examples:

Input: N = 5, r = 2
Output: 10
Explanation: The value of 5C2 is 10

Input: N = 3, r = 1
Output: 3

Approach: The idea is simply based on the below formula.

^NC_r = N! / (r! * (N-r)!)
Also, 
^NC_r-1 = N! / ( (r-1)! * (N – (r-1))! )

Hence,

• ^NC_r * r! * (N – r)! = ^NC_r-1  * (r-1)! * (N – (r-1))!
• ^NC_r * r * (N-r)! = ^NC_r-1  * (N-r+1)!                          [eliminating (r – 1)! from both side]
• ^NC_r * r = ⁿC_r-1  * (N-r+1)

Therefore,

^NC_r = ^NC_r-1 * (N-r+1) / r

Below is the implementation of the above approach:

C++

// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the value of nCr
// using recursion
int nCr(int N, int r)
{
    int res = 0;
    if (r == 0) {
        return 1;
    }
    else {
        res = nCr(N, r - 1)
              * (N - r + 1) / r;
    }
    return res;
}

// Driver code
int main()
{
    int N = 5, r = 3;
    cout << nCr(N, r);
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
class GFG {

  // Function to calculate the value of nCr
  // using recursion
  static int nCr(int N, int r)
  {
    int res = 0;
    if (r == 0) {
      return 1;
    }
    else {
      res = nCr(N, r - 1) * (N - r + 1) / r;
    }
    return res;
  }

  public static void main(String[] args)
  {
    int N = 5, r = 3;

    System.out.println(nCr(N, r));
  }
}

// This code is contributed by Potta Lokesh

Python3

# Python code to implement above approach

# Function to calculate the value Of nCr
# using recursion
def nCr(N, r):
    res = 0
    if(r == 0):
        return 1
    else:
        res = nCr(N, r-1)
* (N-r + 1) / r
    return res


# Driver code
if __name__ == "__main__":
    N = 5
    r = 3
    print(int(nCr(N, r)))

C#

using System;

public class GFG{

  // Function to calculate the value of nCr
  // using recursion
  static int nCr(int N, int r)
  {
    int res = 0;
    if (r == 0) {
      return 1;
    }
    else {
      res = nCr(N, r - 1)
        * (N - r + 1) / r;
    }
    return res;
  }

  // Driver code
  static public void Main (){

    int N = 5, r = 3;
    Console.WriteLine(nCr(N, r));
  }
}

// This code is contributed by hrithikgarg03188.

Javascript

<script>
//Javascript code to implement above approach

// Function to calculate the value of nCr
// using recursion
function nCr(N, r)
{
    let res = 0;
    if (r == 0) {
        return 1;
    }
    else {
        res = nCr(N, r - 1)
              * (N - r + 1) / r;
    }
    return res;
}

// Driver code

let N = 5, r = 3;
document.write(nCr(N,r));

// This code is contributed by Taranpreet
// </script>

Output

10

Time Complexity: O(N!)
Auxiliary Space: O(1)