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Number of ways of distributing N identical objects in R distinct groups
  • Difficulty Level : Easy
  • Last Updated : 07 May, 2021

Given two integers N and R, the task is to calculate the number of ways to distribute N identical objects into R distinct groups.

Examples: 

Input: N = 4, R = 2 
Output:
No of objects in 1st group = 0, in second group = 4 
No of objects in 1st group = 1, in second group = 3 
No of objects in 1st group = 2, in second group = 2 
No of objects in 1st group = 3, in second group = 1 
No of objects in 1st group = 4, in second group = 0

Input: N = 4, R = 3 
Output: 15 

Approach: Idea is to use Multinomial theorem. Let us suppose that x1 objects are placed in the first group, x2 objects are placed in the second group and xR objects are placed in the Rth group. It is given that, 
x1 + x2 + x3 +…+ xR = N 
The solution of this equation by multinomial theorem is given by N + R – 1CR – 1.



Below is the implementation of the above approach: 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the
// value of ncr effectively
int ncr(int n, int r)
{
 
    // Initialize the answer
    int ans = 1;
 
    for (int i = 1; i <= r; i += 1)
    {
        // Divide simultaneously by
        // i to avoid overflow
        ans *= (n - r + i);
        ans /= i;
    }
    return ans;
}
 
// Function to return the number of
// ways to distribute N identical
// objects in R distinct objects
int NoOfDistributions(int N, int R)
{
    return ncr(N + R - 1, R - 1);
}
 
// Driver code
int main()
{
    int N = 4, R = 3;
 
    // Function call
    cout << NoOfDistributions(N, R);
 
    return 0;
}

Java




// Java implementation of the above approach
import java.util.*;
 
class GFG {
 
    // Function to return the
    // value of ncr effectively
    static int ncr(int n, int r)
    {
 
        // Initialize the answer
        int ans = 1;
 
        for (int i = 1; i <= r; i += 1)
        {
            // Divide simultaneously by
            // i to avoid overflow
            ans *= (n - r + i);
            ans /= i;
        }
        return ans;
    }
 
    // Function to return the number of
    // ways to distribute N identical
    // objects in R distinct objects
    static int NoOfDistributions(int N, int R)
    {
        return ncr(N + R - 1, R - 1);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 4, R = 3;
       
        // Function call
        System.out.println(NoOfDistributions(N, R));
    }
}
 
// This code is contributed by Princi Singh

Python3




# Python3 implementation of the above approach
 
# Function to return the
# value of ncr effectively
 
 
def ncr(n, r):
 
    # Initialize the answer
    ans = 1
 
    for i in range(1, r+1):
 
        # Divide simultaneously by
        # i to avoid overflow
        ans *= (n - r + i)
        ans //= i
 
    return ans
 
 
# Function to return the number of
# ways to distribute N identical
# objects in R distinct objects
def NoOfDistributions(N, R):
 
    return ncr(N + R-1, R - 1)
 
 
# Driver code
N = 4
R = 3
 
# Function call
print(NoOfDistributions(N, R))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the above approach
using System;
 
class GFG {
 
    // Function to return the
    // value of ncr effectively
    static int ncr(int n, int r)
    {
 
        // Initialize the answer
        int ans = 1;
 
        for (int i = 1; i <= r; i += 1)
        {
            // Divide simultaneously by
            // i to avoid overflow
            ans *= (n - r + i);
            ans /= i;
        }
        return ans;
    }
 
    // Function to return the number of
    // ways to distribute N identical
    // objects in R distinct objects
    static int NoOfDistributions(int N, int R)
    {
        return ncr(N + R - 1, R - 1);
    }
 
    // Driver code
    static public void Main()
    {
        int N = 4, R = 3;
       
        // Function call
        Console.WriteLine(NoOfDistributions(N, R));
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
 
// Javascript implementation of the above approach
 
// Function to return the
// value of ncr effectively
function ncr(n, r)
{
     
    // Initialize the answer
    let ans = 1;
 
    for(let i = 1; i <= r; i += 1)
    {
         
        // Divide simultaneously by
        // i to avoid overflow
        ans *= (n - r + i);
        ans = parseInt(ans / i);
    }
    return ans;
}
 
// Function to return the number of
// ways to distribute N identical
// objects in R distinct objects
function NoOfDistributions(N, R)
{
    return ncr(N + R - 1, R - 1);
}
 
// Driver code
let N = 4, R = 3;
 
// Function call
document.write(NoOfDistributions(N, R));
    
// This code is contributed by subhammahato348
     
</script>
Output
15

Time Complexity: O(R)

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