# Number of ways of distributing N identical objects in R distinct groups

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: 2
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` ` `  `    ``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``;  ` `            ``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` ` `  `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;  ` `            ``Console.WriteLine(NoOfDistributions(N, R));  ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```15
```

Time Complexity: O(R)

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