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Code Probability

Calculate value of ncr efficiently

The formula to calculate nCr, also known as the combination of n things taken r at a time, is:

nCr = n! / (r! * (n-r)!)

Where "n" is the total number of things, and "r" is the number of things taken at a time.

Let's break down this formula into simpler terms.

Firstly, the exclamation mark, or factorial, means to multiply all the whole numbers from the given number down to 1. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

In the nCr formula, we have to calculate the factorials of both "n" and "r". Then we multiply the factorial of "r" with the factorial of "n-r". Finally, we divide the factorial of "n" with the result of the previous multiplication.

For example, if we want to find the value of 5C2, we will plug in "n=5" and "r=2" into the formula:

5C2 = 5! / (2! * (5-2)!) = 5 x 4 x 3 x 2 x 1 / [(2 x 1) x (3 x 2 x 1)] = 10

Therefore, the value of 5C2 is 10, which means there are 10 ways to choose 2 items out of 5.

In summary, nCr is a formula to calculate the number of ways we can choose "r" items from a total of "n" items. It involves calculating factorials and is a fundamental concept in combinatorics, probability, and statistics.

Code Solution

/*
    Java program
    Calculate value of ncr efficiently
*/
public class Combinations
{
	// Returns the gcd value of two numbers
	public long gcd(long a, long b)
	{
		if (b == 0)
		{
			return a;
		}
		return gcd(b, a % b);
	}
	public void findNcR(int objects, int subset)
	{
		int n = objects;
		int r = subset;
		// Formula of nCr 
		// nCr = n!/[r!(n-r)!]
		// Here
		// C : Indicates number of combination
		// n : Number of objects in set
		// r : Subset size
		long p = 1;
		long k = 1;
		if (n - r < r)
		{
			// Change r when (n - r) is smaller to r
			r = n - r;
		}
		if (r != 0)
		{
			while (r >= 1)
			{
				p = p * n;
				k = k * r;
				// Reduce the value of  n and r by 1
				n--;
				r--;
				// Get gcd of p and k
				long d = gcd(p, k);
				// Divide p and k by d
				// d is common divisor
				p = p / d;
				k = k / d;
			}
		}
		else
		{
			p = 1;
		}
		// Display given value
		System.out.println(" Given n : " + objects + " r : " + subset);
		// Display calculated result
		System.out.println(" Result  : " + p);
	}
	public static void main(String[] args)
	{
		Combinations task = new Combinations();
		// Test Case
		task.findNcR(7, 2);
		task.findNcR(12, 3);
	}
}

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220
// Include header file
#include <iostream>
using namespace std;

/*
    C++ program
    Calculate value of ncr efficiently
*/

class Combinations
{
	public:
		// Returns the gcd value of two numbers
		long gcd(long a, long b)
		{
			if (b == 0)
			{
				return a;
			}
			return this->gcd(b, a % b);
		}
	void findNcR(int objects, int subset)
	{
		int n = objects;
		int r = subset;
		// Formula of nCr 
		// nCr = n!/[r!(n-r)!]
		// Here
		// C : Indicates number of combination
		// n : Number of objects in set
		// r : Subset size
		long p = 1;
		long k = 1;
		if (n - r < r)
		{
			// Change r when (n - r) is smaller to r
			r = n - r;
		}
		if (r != 0)
		{
			while (r >= 1)
			{
				p = p *n;
				k = k *r;
				// Reduce the value of  n and r by 1
				n--;
				r--;
				// Get gcd of p and k
				long d = this->gcd(p, k);
				// Divide p and k by d
				// d is common divisor
				p = p / d;
				k = k / d;
			}
		}
		else
		{
			p = 1;
		}
		// Display given value
		cout << " Given n : " << objects << " r : " << subset << endl;
		// Display calculated result
		cout << " Result  : " << p << endl;
	}
};
int main()
{
	Combinations *task = new Combinations();
	// Test Case
	task->findNcR(7, 2);
	task->findNcR(12, 3);
	return 0;
}

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220
// Include namespace system
using System;
/*
    Csharp program
    Calculate value of ncr efficiently
*/
public class Combinations
{
	// Returns the gcd value of two numbers
	public long gcd(long a, long b)
	{
		if (b == 0)
		{
			return a;
		}
		return this.gcd(b, a % b);
	}
	public void findNcR(int objects, int subset)
	{
		int n = objects;
		int r = subset;
		// Formula of nCr 
		// nCr = n!/[r!(n-r)!]
		// Here
		// C : Indicates number of combination
		// n : Number of objects in set
		// r : Subset size
		long p = 1;
		long k = 1;
		if (n - r < r)
		{
			// Change r when (n - r) is smaller to r
			r = n - r;
		}
		if (r != 0)
		{
			while (r >= 1)
			{
				p = p * n;
				k = k * r;
				// Reduce the value of  n and r by 1
				n--;
				r--;
				// Get gcd of p and k
				long d = this.gcd(p, k);
				// Divide p and k by d
				// d is common divisor
				p = p / d;
				k = k / d;
			}
		}
		else
		{
			p = 1;
		}
		// Display given value
		Console.WriteLine(" Given n : " + objects + " r : " + subset);
		// Display calculated result
		Console.WriteLine(" Result  : " + p);
	}
	public static void Main(String[] args)
	{
		Combinations task = new Combinations();
		// Test Case
		task.findNcR(7, 2);
		task.findNcR(12, 3);
	}
}

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220
<?php
/*
    Php program
    Calculate value of ncr efficiently
*/
class Combinations
{
	// Returns the gcd value of two numbers
	public	function gcd($a, $b)
	{
		if ($b == 0)
		{
			return $a;
		}
		return $this->gcd($b, $a % $b);
	}
	public	function findNcR($objects, $subset)
	{
		$n = $objects;
		$r = $subset;
		// Formula of nCr 
		// nCr = n!/[r!(n-r)!]
		// Here
		// C : Indicates number of combination
		// n : Number of objects in set
		// r : Subset size
		$p = 1;
		$k = 1;
		if ($n - $r < $r)
		{
			// Change r when (n - r) is smaller to r
			$r = $n - $r;
		}
		if ($r != 0)
		{
			while ($r >= 1)
			{
				$p = $p * $n;
				$k = $k * $r;
				// Reduce the value of  n and r by 1
				$n--;
				$r--;
				// Get gcd of p and k
				$d = $this->gcd($p, $k);
				// Divide p and k by d
				// d is common divisor
				$p = $p / $d;
				$k = $k / $d;
			}
		}
		else
		{
			$p = 1;
		}
		// Display given value
		echo(" Given n : ".$objects.
			" r : ".$subset.
			"\n");
		// Display calculated result
		echo(" Result  : ".$p.
			"\n");
	}
}

function main()
{
	$task = new Combinations();
	// Test Case
	$task->findNcR(7, 2);
	$task->findNcR(12, 3);
}
main();

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220
/*
    Node JS program
    Calculate value of ncr efficiently
*/
class Combinations
{
	// Returns the gcd value of two numbers
	gcd(a, b)
	{
		if (b == 0)
		{
			return a;
		}
		return this.gcd(b, a % b);
	}
	findNcR(objects, subset)
	{
		var n = objects;
		var r = subset;
		// Formula of nCr 
		// nCr = n!/[r!(n-r)!]
		// Here
		// C : Indicates number of combination
		// n : Number of objects in set
		// r : Subset size
		var p = 1;
		var k = 1;
		if (n - r < r)
		{
			// Change r when (n - r) is smaller to r
			r = n - r;
		}
		if (r != 0)
		{
			while (r >= 1)
			{
				p = p * n;
				k = k * r;
				// Reduce the value of  n and r by 1
				n--;
				r--;
				// Get gcd of p and k
				var d = this.gcd(p, k);
				// Divide p and k by d
				// d is common divisor
				p = p / d;
				k = k / d;
			}
		}
		else
		{
			p = 1;
		}
		// Display given value
		console.log(" Given n : " + objects + " r : " + subset);
		// Display calculated result
		console.log(" Result  : " + p);
	}
}

function main()
{
	var task = new Combinations();
	// Test Case
	task.findNcR(7, 2);
	task.findNcR(12, 3);
}
main();

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220
#    Python 3 program
#    Calculate value of ncr efficiently
class Combinations :
	#  Returns the gcd value of two numbers
	def gcd(self, a, b) :
		if (b == 0) :
			return a
		
		return self.gcd(b, a % b)
	
	def findNcR(self, objects, subset) :
		n = objects
		r = subset
		#  Formula of nCr 
		#  nCr = n!/[r!(n-r)!]
		#  Here
		#  C : Indicates number of combination
		#  n : Number of objects in set
		#  r : Subset size
		p = 1
		k = 1
		if (n - r < r) :
			#  Change r when (n - r) is smaller to r
			r = n - r
		
		if (r != 0) :
			while (r >= 1) :
				p = p * n
				k = k * r
				#  Reduce the value of  n and r by 1
				n -= 1
				r -= 1
				#  Get gcd of p and k
				d = self.gcd(p, k)
				#  Divide p and k by d
				#  d is common divisor
				p = int(p / d)
				k = int(k / d)
			
		else :
			p = 1
		
		#  Display given value
		print(" Given n : ", objects ," r : ", subset)
		#  Display calculated result
		print(" Result  : ", p)
	

def main() :
	task = Combinations()
	#  Test Case
	task.findNcR(7, 2)
	task.findNcR(12, 3)

if __name__ == "__main__": main()

input

 Given n :  7  r :  2
 Result  :  21
 Given n :  12  r :  3
 Result  :  220
#    Ruby program
#    Calculate value of ncr efficiently
class Combinations 
	#  Returns the gcd value of two numbers
	def gcd(a, b) 
		if (b == 0) 
			return a
		end

		return self.gcd(b, a % b)
	end

	def findNcR(objects, subset) 
		n = objects
		r = subset
		#  Formula of nCr 
		#  nCr = n!/[r!(n-r)!]
		#  Here
		#  C : Indicates number of combination
		#  n : Number of objects in set
		#  r : Subset size
		p = 1
		k = 1
		if (n - r < r) 
			#  Change r when (n - r) is smaller to r
			r = n - r
		end

		if (r != 0) 
			while (r >= 1) 
				p = p * n
				k = k * r
				#  Reduce the value of  n and r by 1
				n -= 1
				r -= 1
				#  Get gcd of p and k
				d = self.gcd(p, k)
				#  Divide p and k by d
				#  d is common divisor
				p = p / d
				k = k / d
			end

		else
 
			p = 1
		end

		#  Display given value
		print(" Given n : ", objects ," r : ", subset, "\n")
		#  Display calculated result
		print(" Result  : ", p, "\n")
	end

end

def main() 
	task = Combinations.new()
	#  Test Case
	task.findNcR(7, 2)
	task.findNcR(12, 3)
end

main()

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220

/*
    Scala program
    Calculate value of ncr efficiently
*/
class Combinations()
{
	// Returns the gcd value of two numbers
	def gcd(a: Long, b: Long): Long = {
		if (b == 0)
		{
			return a;
		}
		return gcd(b, a % b);
	}
	def findNcR(objects: Int, subset: Int): Unit = {
		var n: Int = objects;
		var r: Int = subset;
		// Formula of nCr 
		// nCr = n!/[r!(n-r)!]
		// Here
		// C : Indicates number of combination
		// n : Number of objects in set
		// r : Subset size
		var p: Long = 1;
		var k: Long = 1;
		if (n - r < r)
		{
			// Change r when (n - r) is smaller to r
			r = n - r;
		}
		if (r != 0)
		{
			while (r >= 1)
			{
				p = p * n;
				k = k * r;
				// Reduce the value of  n and r by 1
				n -= 1;
				r -= 1;
				// Get gcd of p and k
				var d: Long = gcd(p, k);
				// Divide p and k by d
				// d is common divisor
				p = p / d;
				k = k / d;
			}
		}
		else
		{
			p = 1;
		}
		// Display given value
		println(" Given n : " + objects + " r : " + subset);
		// Display calculated result
		println(" Result  : " + p);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Combinations = new Combinations();
		// Test Case
		task.findNcR(7, 2);
		task.findNcR(12, 3);
	}
}

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220
/*
    Swift 4 program
    Calculate value of ncr efficiently
*/
class Combinations
{
	// Returns the gcd value of two numbers
	func gcd(_ a: Int, _ b: Int) -> Int
	{
		if (b == 0)
		{
			return a;
		}
		return self.gcd(b, a % b);
	}
	func findNcR(_ objects: Int, _ subset: Int)
	{
		var n: Int = objects;
		var r: Int = subset;
		// Formula of nCr 
		// nCr = n!/[r!(n-r)!]
		// Here
		// C : Indicates number of combination
		// n : Number of objects in set
		// r : Subset size
		var p: Int = 1;
		var k: Int = 1;
		if (n - r < r)
		{
			// Change r when (n - r) is smaller to r
			r = n - r;
		}
		if (r  != 0)
		{
			while (r >= 1)
			{
				p = p * n;
				k = k * r;
				// Reduce the value of  n and r by 1
				n -= 1;
				r -= 1;
				// Get gcd of p and k
				let d: Int = self.gcd(p, k);
				// Divide p and k by d
				// d is common divisor
				p = p / d;
				k = k / d;
			}
		}
		else
		{
			p = 1;
		}
		// Display given value
		print(" Given n : ", objects ," r : ", subset);
		// Display calculated result
		print(" Result  : ", p);
	}
}
func main()
{
	let task: Combinations = Combinations();
	// Test Case
	task.findNcR(7, 2);
	task.findNcR(12, 3);
}
main();

input

 Given n :  7  r :  2
 Result  :  21
 Given n :  12  r :  3
 Result  :  220
/*
    Kotlin program
    Calculate value of ncr efficiently
*/
class Combinations
{
	// Returns the gcd value of two numbers
	fun gcd(a: Long, b: Long): Long
	{
		if (b == 0L)
		{
			return a;
		}
		return this.gcd(b, a % b);
	}
	fun findNcR(objects: Int, subset: Int): Unit
	{
		var n: Int = objects;
		var r: Int = subset;
		// Formula of nCr 
		// nCr = n!/[r!(n-r)!]
		// Here
		// C : Indicates number of combination
		// n : Number of objects in set
		// r : Subset size
		var p: Long = 1;
		var k: Long = 1;
		if (n - r < r)
		{
			// Change r when (n - r) is smaller to r
			r = n - r;
		}
		if (r != 0)
		{
			while (r >= 1)
			{
				p = p * n;
				k = k * r;
				// Reduce the value of  n and r by 1
				n -= 1;
				r -= 1;
				// Get gcd of p and k
				val d: Long = this.gcd(p, k);
				// Divide p and k by d
				// d is common divisor
				p = p / d;
				k = k / d;
			}
		}
		else
		{
			p = 1;
		}
		// Display given value
		println(" Given n : " + objects + " r : " + subset);
		// Display calculated result
		println(" Result  : " + p);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Combinations = Combinations();
	// Test Case
	task.findNcR(7, 2);
	task.findNcR(12, 3);
}

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220
package main
import "fmt"
/*
    Go program
    Calculate value of ncr efficiently
*/

// Returns the gcd value of two numbers
func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a % b)
}
func findNcR(objects, subset int) {
	var n int = objects
	var r int = subset
	// Formula of nCr 
	// nCr = n!/[r!(n-r)!]
	// Here
	// C : Indicates number of combination
	// n : Number of objects in set
	// r : Subset size
	var p int = 1
	var k int = 1
	if n - r < r {
		// Change r when (n - r) is smaller to r
		r = n - r
	}
	if r != 0 {
		for (r >= 1) {
			p = p * n
			k = k * r
			// Reduce the value of  n and r by 1
			n--
			r--
			// Get gcd of p and k
			var d int =  gcd(p, k)
			// Divide p and k by d
			// d is common divisor
			p = p / d
			k = k / d
		}
	} else {
		p = 1
	}
	// Display given value
	fmt.Println(" Given n : ", objects, " r : ", subset)
	// Display calculated result
	fmt.Println(" Result  : ", p)
}
func main() {
	// Test Case
	findNcR(7, 2)
	findNcR(12, 3)
}

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220
/*
    Rust program
    Calculate value of ncr efficiently
*/
fn main(){
	// Test Case
	find_n_c_r(7, 2);
	find_n_c_r(12, 3);
}
// Returns the gcd value of two numbers
fn  gcd( a: i32, b: i32) -> i32
{
	if b == 0
	{
		return a;
	}
	return gcd(b, a % b);
}
fn find_n_c_r( objects: i32, subset: i32)
{
	let mut n: i32 = objects;
	let mut r: i32 = subset;
	// Formula of nCr 
	// nCr = n!/[r!(n-r)!]
	// Here
	// C : Indicates number of combination
	// n : Number of objects in set
	// r : Subset size
	let mut p = 1;
	let mut k = 1;
	if (n - r) < r
	{
		// Change r when (n - r) is smaller to r
		r = n - r;
	}
	if r != 0
	{
		while r >= 1
		{
			p = p * n;
			k = k * r;
			// Reduce the value of  n and r by 1
			n=n-1;
			r=r-1;
			// Get gcd of p and k
			let d =  gcd(p, k);
			// Divide p and k by d
			// d is common divisor
			p = p / d;
			k = k / d;
		}
	}
	else
	{
		p = 1;
	}
	// Display given value
	println!(" Given n : {0} r : {1} ", objects , subset);
	// Display calculated result
	println!(" Result  : {0}" , p);
}

input

 Given n : 7 r : 2
 Result  : 21
 Given n : 12 r : 3
 Result  : 220

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