Number of ways of distributing N identical objects in R distinct groups with no groups empty - GeeksforGeeks

18 Jan, 2022
Given two integer N and R, the task is to calculate the number of ways to distribute N identical objects into R distinct groups such that no groups are left empty.
Examples: 
Input: N = 4, R = 2 Output: 3 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
Input: N = 5, R = 3 Output: 6  
Approach: Idea is to use Multinomial theorem. Let us suppose that x1 objects are placed in the first group, x2 objects are placed in second group and xR objects are placed in the Rth group. It is given that, x1 + x2 + x3 +...+ xR = N for all xi ≥ 1 for 1 ≤ i ≤ R Now replace every xi with yi + 1 for all 1 ≤ i ≤ R. Now all the y variables are greater than or equal to zero. The equation becomes, y1 + y2 + y3 + ... + yR + R = N for all yi ≥ 0 for 1 ≤ i ≤ R y1 + y2 + y3 + ... + yR = N – R It now reduces to that standard multinomial equation whose solution is (N – R) + R – 1CR – 1. The solution of this equation is given by N – 1CR – 1.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std; // Function to return the// value of ncr effectivelyint 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 objectsint NoOfDistributions(int N, int R){    return ncr(N - 1, R - 1);} // Driver codeint main(){    int N = 4;    int R = 3;     cout << NoOfDistributions(N, R);     return 0;}
// Java implementation of the above approachimport java.io.*; 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 - 1, R - 1);    }         // Driver code    public static void main (String[] args)    {         int N = 4;        int R = 3;             System.out.println(NoOfDistributions(N, R));    }} // This code is contributed by ajit
# Python3 implementation of the above approach # Function to return the# value of ncr effectivelydef 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 objectsdef NoOfDistributions(N, R):     return ncr(N - 1, R - 1) # Driver code N = 4R = 3 print(NoOfDistributions(N, R)) # This code is contributed by mohit kumar 29
// C# implementation of the above approachusing 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 - 1, R - 1);    }         // Driver code    static public void Main ()    {        int N = 4;        int R = 3;             Console.WriteLine(NoOfDistributions(N, R));    }} // This code is contributed by AnkitRai01
<script> // Javascript implementation of the above approach // Function to return the// value of ncr effectivelyfunction 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 objectsfunction NoOfDistributions(N, R){    return ncr(N - 1, R - 1);} // Driver codelet N = 4;let R = 3; document.write(NoOfDistributions(N, R)); // This code is contributed by rishavmahato348 </script>
3
 
Time Complexity: O(R) 
mohit kumar 29
ankthon
jit_t
rishavmahato348
varshagumber28
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