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: 5
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 = 0Input: 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> |
15
Time Complexity: O(R)
Auxiliary Space: O(1)
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