# 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:
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 particular 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 ` `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

 ``

Output:

`10`

Time Complexity: O(X)

Auxiliary Space: O(1)

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