Number of ways of distributing N identical objects in R distinct groups with no groups empty
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 variaables 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:
CPP
// 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 - 1, R - 1); } // Driver code int main() { int N = 4; int R = 3; cout << NoOfDistributions(N, R); return 0; } |
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Java
// Java implementation of the above approach import 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 |
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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 - 1, R - 1) # Driver code N = 4R = 3 print(NoOfDistributions(N, R)) # This code is contributed by mohit kumar 29 |
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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 - 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 |
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Output:
3
Time Complexity: O(R)
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