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Showing content from https://www.geeksforgeeks.org/dsa/program-to-calculate-value-of-ncr-using-recursion/ below:

Program to calculate value of nCr using Recursion

Last Updated : 23 Jul, 2025

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 1: One very interesting way to find the C(n,r) is the recursive method which is based on the recursive equation.

C(n,r) = C(n-1,r-1) + C(n-1,r)

Below is the implementation of the above approach as follows:

C++


 #include <bits/stdc++.h>
using namespace std;

// Function to calculate the value of nCr
// using recursion

int comb(int n,int r){
   if(n<r){
       return 0;
   }
   if(r == 0){
       return 1;
   }
   if(r == 1){
       return n;
   }
   if(n == 1){
       return 1;
   }
   return comb(n-1,r-1)+comb(n-1,r);
}

// Driver code
int main(){
   int n = 10,r = 5;
   cout << comb(n,r);
}

Java


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

  // Function to calculate the value of nCr
  // using recursion
  static int comb(int n, int r)
  {
     if(n<r){
       return 0;
     }
     if(r == 0){
         return 1;
     }
     if(r == 1){
         return n;
     }
     if(n == 1){
         return 1;
     }
     return comb(n-1,r-1)+comb(n-1,r);
  }

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

    System.out.println(comb(n, r));
  }
}

Python3


 # Python code to implement above approach

# Function to calculate the value of nCr
# using recursion

def comb(n, r):
    if(n < r):
        return 0
    if(r == 0):
        return 1
    if(r == 1):
        return n
    if(n == 1):
        return 1
    return comb(n - 1, r - 1) + comb(n - 1, r)
    
#  Driver code
n = 10
r = 5
print(comb(n, r))

# This code is contributed by Pushpesh Raj.


 using System;

public class GFG{

  // Function to calculate the value of nCr
  // using recursion
  static int comb(int n, int r)
  {
    if(n<r){
       return 0;
    }
    if(r == 0){
       return 1;
    }
    if(r == 1){
       return n;
    }
    if(n == 1){
       return 1;
    }
    return comb(n-1,r-1)+comb(n-1,r);
  }

  // Driver code
  static public void Main (){

    int n = 5, r = 3;
    Console.WriteLine(comb(n, r));
  }
}

JavaScript


 <script>
//Javascript code to implement above approach

// Function to calculate the value of nCr
// using recursion

function comb(n, r){
   if( n<r ){
       return 0;
   }
   if(r == 0){
       return 1;
   }
   if(r == 1){
       return n;
   }
   if(n == 1){
       return 1;
   }
   return comb(n-1,r-1) + comb(n-1,r);
}

// Driver code

let n = 5, r = 3;
document.write(comb(n,r));
 
// </script>

Time Complexity: O(2^n), Auxiliary Space: O(n)

Approach 2: Another 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)))


 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>

Time Complexity: O(r), Auxiliary Space: O(r)

Complexity Analysis:
The time complexity of the above approach is O(r). This is because the function makes a single recursive call for each value of r, and the time taken to calculate the value of nCr is constant.

The Auxiliary space complexity of the above approach is O(r), as the recursive function calls create a new stack frame for each call. This means that the program will consume a significant amount of memory for larger values of r.

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