Total ways of selecting a group of X men from N men with or without including a particular man

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 ^NC_X

• Including a particular man: We can choose (X – 1) men from (N – 1) in ^{N – 1}C_{X – 1}.
• Excluding a particular man: We can choose X men from (N – 1) in ^{N – 1}C_X

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

Time Complexity: O(X)

Auxiliary Space: O(1)