Program to calculate the value of nCr Efficiently
Last Updated :
11 Jul, 2025
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Given two numbers n, r ( n>=r ). The task is to find the value of C(n, r) for big value of n.
Examples:
Input: n = 30, r = 15 Output: 155117520 C(30, 15) is 155117520 by 30!/((30-15)!*15!) Input: n = 50, r = 25 Output: 126410606437752
Approach: A simple code can be created with the following knowledge that :
C(n, r) = [n * (n-1) * .... * (n-r+1)] / [r * (r-1) * .... * 1]
However, for big values of n, r the products may overflow, hence during each iteration we divide the current variables holding value of products by their gcd.
Below is the required implementation:
// C++ implementation to find nCr
#include <bits/stdc++.h>
using namespace std;
// Function to find the nCr
void printNcR(int n, int r)
{
// p holds the value of n*(n-1)*(n-2)...,
// k holds the value of r*(r-1)...
long long p = 1, k = 1;
// C(n, r) == C(n, n-r),
// choosing the smaller value
if (n - r < r)
r = n - r;
if (r != 0) {
while (r) {
p *= n;
k *= r;
// gcd of p, k
long long m = __gcd(p, k);
// dividing by gcd, to simplify
// product division by their gcd
// saves from the overflow
p /= m;
k /= m;
n--;
r--;
}
// k should be simplified to 1
// as C(n, r) is a natural number
// (denominator should be 1 ) .
}
else
p = 1;
// if our approach is correct p = ans and k =1
cout << p << endl;
}
// Driver code
int main()
{
int n = 50, r = 25;
printNcR(n, r);
return 0;
}
// Java implementation to find nCr
class GFG {
// Function to find the nCr
static void printNcR(int n, int r)
{
// p holds the value of n*(n-1)*(n-2)...,
// k holds the value of r*(r-1)...
long p = 1, k = 1;
// C(n, r) == C(n, n-r),
// choosing the smaller value
if (n - r < r) {
r = n - r;
}
if (r != 0) {
while (r > 0) {
p *= n;
k *= r;
// gcd of p, k
long m = __gcd(p, k);
// dividing by gcd, to simplify
// product division by their gcd
// saves from the overflow
p /= m;
k /= m;
n--;
r--;
}
// k should be simplified to 1
// as C(n, r) is a natural number
// (denominator should be 1 ) .
}
else {
p = 1;
}
// if our approach is correct p = ans and k =1
System.out.println(p);
}
static long __gcd(long n1, long n2)
{
long gcd = 1;
for (int i = 1; i <= n1 && i <= n2; ++i) {
// Checks if i is factor of both integers
if (n1 % i == 0 && n2 % i == 0) {
gcd = i;
}
}
return gcd;
}
// Driver code
public static void main(String[] args)
{
int n = 50, r = 25;
printNcR(n, r);
}
}
# Python3 implementation to find nCr
from math import *
# Function to find the nCr
def printNcR(n, r):
# p holds the value of n*(n-1)*(n-2)...,
# k holds the value of r*(r-1)...
p = 1
k = 1
# C(n, r) == C(n, n-r),
# choosing the smaller value
if (n - r < r):
r = n - r
if (r != 0):
while (r):
p *= n
k *= r
# gcd of p, k
m = gcd(p, k)
# dividing by gcd, to simplify product
# division by their gcd saves from
# the overflow
p //= m
k //= m
n -= 1
r -= 1
# k should be simplified to 1
# as C(n, r) is a natural number
# (denominator should be 1 )
else:
p = 1
# if our approach is correct p = ans and k =1
print(p)
# Driver code
if __name__ == "__main__":
n = 50
r = 25
printNcR(n, r)
# this code is contributed by
# ChitraNayal
// C# implementation to find nCr
using System;
public class GFG {
// Function to find the nCr
static void printNcR(int n, int r)
{
// p holds the value of n*(n-1)*(n-2)...,
// k holds the value of r*(r-1)...
long p = 1, k = 1;
// C(n, r) == C(n, n-r),
// choosing the smaller value
if (n - r < r) {
r = n - r;
}
if (r != 0) {
while (r > 0) {
p *= n;
k *= r;
// gcd of p, k
long m = __gcd(p, k);
// dividing by gcd, to simplify
// product division by their gcd
// saves from the overflow
p /= m;
k /= m;
n--;
r--;
}
// k should be simplified to 1
// as C(n, r) is a natural number
// (denominator should be 1 ) .
}
else {
p = 1;
}
// if our approach is correct p = ans and k =1
Console.WriteLine(p);
}
static long __gcd(long n1, long n2)
{
long gcd = 1;
for (int i = 1; i <= n1 && i <= n2; ++i) {
// Checks if i is factor of both integers
if (n1 % i == 0 && n2 % i == 0) {
gcd = i;
}
}
return gcd;
}
// Driver code
static public void Main()
{
int n = 50, r = 25;
printNcR(n, r);
}
// This code is contributed by ajit.
}
<?php
// PHP implementation to find nCr
// Function to find the nCr
function printNcR($n, $r) {
// p holds the value of n*(n-1)*(n-2)...,
// k holds the value of r*(r-1)...
$p = 1;
$k = 1;
// C(n, r) == C(n, n-r),
// choosing the smaller value
if ($n - $r < $r) {
$r = $n - $r;
}
if ($r != 0) {
while ($r > 0) {
$p *= $n;
$k *= $r;
// gcd of p, k
$m = __gcd($p, $k);
// dividing by gcd, to simplify product
// division by their gcd saves from the overflow
$p /= $m;
$k /= $m;
$n--;
$r--;
}
// k should be simplified to 1
// as C(n, r) is a natural number
// (denominator should be 1 ) .
} else {
$p = 1;
}
// if our approach is correct p = ans and k =1
echo ($p);
}
function __gcd($n1, $n2) {
$gcd = 1;
for ($i = 1; $i <= $n1 && $i <= $n2; ++$i) {
// Checks if i is factor of both integers
if ($n1 % $i == 0 && $n2 % $i == 0) {
$gcd = $i;
}
}
return $gcd;
}
// Driver code
$n = 50;
$r = 25;
printNcR($n, $r);
//This code is contributed by Sachin.
?>
<script>
// Javascript implementation to find nCr
function __gcd(n1, n2)
{
var gcd = 1;
for (var i = 1; i <= n1 && i <= n2; ++i) {
// Checks if i is factor of both integers
if (n1 % i == 0 && n2 % i == 0) {
gcd = i;
}
}
return gcd;
}
// Function to find the nCr
function printNcR(n, r)
{
// p holds the value of n*(n-1)*(n-2)...,
// k holds the value of r*(r-1)...
var p = 1, k = 1;
// C(n, r) == C(n, n-r),
// choosing the smaller value
if (n - r < r)
r = n - r;
if (r != 0) {
while (r) {
p *= n;
k *= r;
// gcd of p, k
var m = __gcd(p, k);
// dividing by gcd, to simplify
// product division by their gcd
// saves from the overflow
p /= m;
k /= m;
n--;
r--;
}
// k should be simplified to 1
// as C(n, r) is a natural number
// (denominator should be 1 ) .
}
else
p = 1;
// if our approach is correct p = ans and k =1
document.write(p);
}
// Driver code
var n = 50, r = 25;
printNcR(n, r);
</script>
Output
126410606437752
Time Complexity: O( R Log N)
Auxiliary Space: O(1), since no extra space has been taken.
