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Program to calculate value of nCr

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  • Difficulty Level : Easy
  • Last Updated : 03 Feb, 2023
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Following are the common definitions of Binomial Coefficients

  • A binomial coefficient C(n, k) can be defined as the coefficient of Xk in the expansion of (1 + X)n.
  • A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects; more formally, the number of k-element subsets (or k-combinations) of an n-element set.

Given two numbers N and r, The task is to find the value of NCr

Examples : 

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

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

Approach: Below is the idea to solve the problem:

The total number of ways for selecting r elements out of n options are nCr = (n!) / (r! * (n-r)!) 
where n! = 1 * 2 * . . . * n.

Below is the Implementation of the above approach:

C++




// CPP program To calculate The Value Of nCr
#include <bits/stdc++.h>
using namespace std;
 
int fact(int n);
 
int nCr(int n, int r)
{
    return fact(n) / (fact(r) * fact(n - r));
}
 
// Returns factorial of n
int fact(int n)
{
      if(n==0)
      return 1;
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
 
// Driver code
int main()
{
    int n = 5, r = 3;
    cout << nCr(n, r);
    return 0;
}

C




#include <stdio.h>
 
int factorial(int n) {
      if(n == 0)
      return 1;
    int factorial = 1;
    for (int i = 2; i <= n; i++)
        factorial = factorial * i;
    return factorial;
}
 
int nCr(int n, int r) {
    return factorial(n) / (factorial(r) * factorial(n - r));
}
 
int main() {
    int n = 5, r = 3;
      printf("%d", nCr(n, r));
    return 0;
}
 
// This code was contributed by Omkar Prabhune

Java




// Java program To calculate
// The Value Of nCr
import java.io.*;
 
public class GFG {
 
static int nCr(int n, int r)
{
    return fact(n) / (fact(r) *
                  fact(n - r));
}
 
// Returns factorial of n
static int fact(int n)
{
      if(n==0)
      return 1;
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 5, r = 3;
    System.out.println(nCr(n, r));
}
}
 
// This code is Contributed by
// Smitha Dinesh Semwal.

Python 3




# Python 3 program To calculate
# The Value Of nCr
 
def nCr(n, r):
 
    return (fact(n) / (fact(r)
                * fact(n - r)))
 
# Returns factorial of n
def fact(n):
    if n == 0:
        return 1
    res = 1
     
    for i in range(2, n+1):
        res = res * i
         
    return res
 
# Driver code
n = 5
r = 3
print(int(nCr(n, r)))
 
# This code is contributed
# by Smitha

C#




// C# program To calculate
// The Value Of nCr
using System;
 
class GFG {
 
static int nCr(int n, int r)
{
   return fact(n) / (fact(r) *
                 fact(n - r));
}
 
// Returns factorial of n
static int fact(int n)
{
      if(n==0)
      return 1;
    int res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}
 
   // Driver code
   public static void Main()
   {
      int n = 5, r = 3;
      Console.Write(nCr(n, r));
   }
}
 
// This code is Contributed by nitin mittal.

PHP




<?php
// PHP program To calculate
// the Value Of nCr
 
 
function nCr( $n, $r)
{
    return fact($n) / (fact($r) *
                  fact($n - $r));
}
 
// Returns factorial of n
function fact( $n)
{
      if($n == 0)
      return 1;
    $res = 1;
    for ( $i = 2; $i <= $n; $i++)
        $res = $res * $i;
    return $res;
}
 
    // Driver code
    $n = 5;
    $r = 3;
    echo nCr($n, $r);
     
// This code is contributed by vt_m.
?>

Javascript




<script>
 
// Javascript program To calculate The Value Of nCr
 
function nCr(n, r)
{
    return fact(n) / (fact(r) * fact(n - r));
}
 
// Returns factorial of n
function fact(n)
{
      if(n==0)
      return 1;
    var res = 1;
    for (var i = 2; i <= n; i++)
        res = res * i;
    return res;
}
 
// Driver code
var n = 5, r = 3;
document.write(nCr(n, r));
 
 
</script>

Output

10

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

The time complexity of the above approach is O(N).
This is because the function fact() has a time complexity of O(N), and it is called twice for each call to nCr().

The space complexity of the above approach is O(1).
Because the function does not make any recursive calls and only uses a constant amount of memory.

Another Approach:

The idea is to use a recursive function to calculate the value of nCr. The base cases are:

  • if r is greater than n, return 0 (there are no combinations possible)
  • if r is 0 or r is n, return 1 (there is only 1 combination possible in these cases)

For other values of n and r, the function calculates the value of nCr by adding the number of combinations possible by including the current element and the number of combinations possible by not including the current element.

Below is the Implementation of the above approach:
 

C++




#include <iostream>
using namespace std;
 
int nCr(int n, int r)
{
    if (r > n)
        return 0;
    if (r == 0 || r == n)
        return 1;
    return nCr(n - 1, r - 1) + nCr(n - 1, r);
}
 
int main()
{
 
    cout << nCr(5, 3); // Output: 10
    return 0;
}
 
// This code is contributed by Susobhan Akhuli

Java




import java.util.*;
 
class GFG {
    public static int nCr(int n, int r)
    {
        if (r > n)
            return 0;
        if (r == 0 || r == n)
            return 1;
        return nCr(n - 1, r - 1) + nCr(n - 1, r);
    }
 
    public static void main(String[] args)
    {
        System.out.println(nCr(5, 3)); // Output: 10
    }
}
 
// This code is contributed by Prasad Kandekar(prasad264)

Python3




def nCr(n, r):
    if r > n:
        return 0
    if r == 0 or r == n:
        return 1
    return nCr(n-1, r-1) + nCr(n-1, r)
 
 
print(nCr(5, 3))  # Output: 10
 
# This code is contributed by Susobhan Akhuli

C#




using System;
 
public class GFG {
    static public int nCr(int n, int r)
    {
        if (r > n)
            return 0;
        if (r == 0 || r == n)
            return 1;
        return nCr(n - 1, r - 1) + nCr(n - 1, r);
    }
 
    static public void Main(string[] args)
    {
        Console.WriteLine(nCr(5, 3)); // Output: 10
    }
}
 
// This code is contributed by Prasad Kandekar(prasad264)

Javascript




function nCr(n, r) {
    if (r > n)
        return 0;
    if (r === 0 || r === n)
        return 1;
    return nCr(n-1, r-1) + nCr(n-1, r);
}
 
console.log(nCr(5, 3)); // Output: 10
 
// This code is contributed by Prasad Kandekar(prasad264)

Output

10

Time Complexity: O(2N)
Auxiliary Space: O(N2)

More Efficient Solutions: 
Dynamic Programming | Set 9 (Binomial Coefficient) 
Space and time efficient Binomial Coefficient 
All Articles on Binomial Coefficient


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