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 nCrint 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 waysint total_ways(int N, int X){ return (nCr(N - 1, X - 1) + nCr(N - 1, X));}// Driver codeint main(){ int N = 5, X = 3; cout << total_ways(N, X); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG{ // Function to return the value of nCrstatic 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 waysstatic int total_ways(int N, int X){ return (nCr(N - 1, X - 1) + nCr(N - 1, X));}// Driver codepublic 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 nCrdef 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 waysdef total_ways(N, X) : return (nCr(N - 1, X - 1) + nCr(N - 1, X));# Driver codeif __name__ == "__main__" : N = 5; X = 3; print(total_ways(N, X));# This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG{ // Function to return the value of nCrstatic 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 waysstatic int total_ways(int N, int X){ return (nCr(N - 1, X - 1) + nCr(N - 1, X));}// Driver codepublic static void Main (String[] args){ int N = 5, X = 3; Console.WriteLine(total_ways(N, X));}} // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation of the approach // Function to return the value of nCr 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 /= i; } return ans; } // Function to return the count of ways function total_ways(N, X) { return (nCr(N - 1, X - 1) + nCr(N - 1, X)); } let N = 5, X = 3; document.write(total_ways(N, X)); </script> |
10
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