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Total ways of selecting a group of X men from N men with or without including a particular man
  • Last Updated : 10 Mar, 2021

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:
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

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>
Output: 
10

 

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