Given two integers X and N. The task is to find the total number of ways of selecting X men from a group of N men with or without including a particular man.
Examples:
Input: N = 3 X = 2
Output: 3
Including a man say M1, the ways can be (M1, M2) and (M1, M3).
Excluding a man say M1, the only way is (M2, M3).
Total ways = 2 + 1 = 3.Input: N = 5 X = 3
Output: 10
Approach: The total number of ways of choosing X men from N men is NCX
- Including a particluar man: We can choose (X – 1) men from (N – 1) in N – 1CX – 1.
- Excluding a particular man: We can choose X men from (N – 1) in N – 1CX
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the value of nCr 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 count of ways int total_ways(int N, int X) { return (nCr(N - 1, X - 1) + nCr(N - 1, X)); } // Driver code int main() { int N = 5, X = 3; cout << total_ways(N, X); return 0; } |
Java
// Java implementation of the approach import java.io.*; class GFG { // Function to return the value of nCr 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 count of ways static int total_ways(int N, int X) { return (nCr(N - 1, X - 1) + nCr(N - 1, X)); } // Driver code public static void main (String[] args) { int N = 5, X = 3; System.out.println (total_ways(N, X)); } } // This code is contributed by Sachin |
Python3
# Python3 implementation of the approach # Function to return the value of nCr 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 count of ways def total_ways(N, X) : return (nCr(N - 1, X - 1) + nCr(N - 1, X)); # Driver code if __name__ == "__main__" : N = 5; X = 3; print(total_ways(N, X)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System; class GFG { // Function to return the value of nCr 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 count of ways static int total_ways(int N, int X) { return (nCr(N - 1, X - 1) + nCr(N - 1, X)); } // Driver code public static void Main (String[] args) { int N = 5, X = 3; Console.WriteLine(total_ways(N, X)); } } // This code is contributed by 29AjayKumar |
10
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