Count of groups among N people having only one leader in each group

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Given N number of people, the task is to count the number of ways to form groups of size? N where, in each group, the first element of the group is the leader of the group.
Note:

Groups with same people having different leaders are treated as a different group. For Example: The group {1, 2, 3} and {2, 1, 3} are treated as different group as they have different leader 1 and 2 respectively.
Groups with same leader having same people are treated as a same group. For Example: The groups {1, 3, 2} and {1, 2, 3} are treated as same group as they have same leader and same people.
The answer can be very large, take modulo to (1e9+7).

Examples:

Input: N = 3
Output: 12
Explanation:
Total Groups with leaders are:
Groups with Leader 1:
1. {1}
2. {1, 2}
3. {1, 3}
4. {1, 2, 3}
Groups with Leader 2:
5. {2}
6. {2, 1}
7. {2, 3}
8. {2, 1, 3}
Groups with Leader 3:
9. {3}
10. {3, 1}
11. {3, 2}
12. {3, 1, 2}
Input: N = 5
Output: 80

Approach: This problem can be solved using the concept of Binomial coefficients and modular exponentiation. Below are the observations to this problem statement:

The number of ways to select one leader among N persons is C(N, 1).
For every leader we can select a group of size K where 0 ? K ? N-1 to make the possible number of grouping.
So the total number ways is given by the product of N and the summation of selection K elements from the remaining (N – 1) elements as:

Total Ways = $\binom{N}{1} * \left ( \binom{N-1}{0} + \binom{N-1}{1} + \binom{N-1}{2} + ... + \binom{N-1}{N-1} \right )$

By using Binomial Theorem, the summation of the Binomial Coefficient can be written as:

$\binom{N}{1} * \left ( \binom{N-1}{0} + \binom{N-1}{1} + \binom{N-1}{2} + ... + \binom{N-1}{N-1} \right ) = 2^{N-1}$

Therefore the number of ways of selecting groups having only one leader is

$N * 2^{N - 1}$

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
long long mod = 1000000007;
 
// Function to find 2^x using
// modular exponentiation
int exponentMod(int A, int B)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
 
    // If B is even
    long long y;
    if (B % 2 == 0) {
        y = exponentMod(A, B / 2);
        y = (y * y) % mod;
    }
 
    // If B is odd
    else {
        y = A % mod;
        y = (y * exponentMod(A, B - 1)
             % mod)
            % mod;
    }
 
    return (int)((y + mod) % mod);
}
 
// Function to count the number of
// ways to form the group having
// one leader
void countWays(int N)
{
 
    // Find 2^(N-1) using modular
    // exponentiation
    long long select = exponentMod(2,
                                   N - 1);
 
    // Count total ways
    long long ways
        = ((N % mod)
           * (select % mod));
 
    ways %= mod;
 
    // Print the total ways
    cout << ways;
}
 
// Driver Code
int main()
{
 
    // Given N number of peoples
    int N = 5;
 
    // Function Call
    countWays(N);
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
static long mod = 1000000007;
 
// Function to find 2^x using
// modular exponentiation
static int exponentMod(int A, int B)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
 
    // If B is even
    long y;
    if (B % 2 == 0) 
    {
        y = exponentMod(A, B / 2);
        y = (y * y) % mod;
    }
 
    // If B is odd
    else
    {
        y = A % mod;
        y = (y * exponentMod(A, B - 1) % 
                                  mod) % mod;
    }
 
    return (int)((y + mod) % mod);
}
 
// Function to count the number of
// ways to form the group having
// one leader
static void countWays(int N)
{
 
    // Find 2^(N-1) using modular
    // exponentiation
    long select = exponentMod(2, N - 1);
 
    // Count total ways
    long ways = ((N % mod) * (select % mod));
 
    ways %= mod;
 
    // Print the total ways
    System.out.print(ways);
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given N number of peoples
    int N = 5;
 
    // Function Call
    countWays(N);
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
mod = 1000000007
 
# Function to find 2^x using
# modular exponentiation
def exponentMod(A, B):
     
    # Base cases
    if (A == 0):
        return 0;
    if (B == 0):
        return 1;
 
    # If B is even
    y = 0;
     
    if (B % 2 == 0):
        y = exponentMod(A, B // 2);
        y = (y * y) % mod;
 
    # If B is odd
    else:
        y = A % mod;
        y = (y * exponentMod(A, B - 1) %
                                  mod) % mod;
                              
    return ((y + mod) % mod);
 
# Function to count the number of
# ways to form the group having
# one leader
def countWays(N):
     
    # Find 2^(N-1) using modular
    # exponentiation
    select = exponentMod(2, N - 1);
 
    # Count total ways
    ways = ((N % mod) * (select % mod));
 
    ways %= mod;
 
    # Print the total ways
    print(ways)
     
# Driver code        
if __name__=='__main__':
     
    # Given N number of people
    N = 5;
 
    # Function call
    countWays(N);
 
# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
class GFG{
  
static long mod = 1000000007;
  
// Function to find 2^x using
// modular exponentiation
static int exponentMod(int A, int B)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
  
    // If B is even
    long y;
    if (B % 2 == 0) 
    {
        y = exponentMod(A, B / 2);
        y = (y * y) % mod;
    }
  
    // If B is odd
    else
    {
        y = A % mod;
        y = (y * exponentMod(A, B - 1) % 
                                  mod) % mod;
    }
  
    return (int)((y + mod) % mod);
}
  
// Function to count the number of
// ways to form the group having
// one leader
static void countWays(int N)
{
  
    // Find 2^(N-1) using modular
    // exponentiation
    long select = exponentMod(2, N - 1);
  
    // Count total ways
    long ways = ((N % mod) * (select % mod));
  
    ways %= mod;
  
    // Print the total ways
    Console.Write(ways);
}
  
// Driver Code
public static void Main(String[] args)
{
  
    // Given N number of peoples
    int N = 5;
  
    // Function Call
    countWays(N);
}
}
 
// This code is contributed by sapnasingh4991

Javascript

<script>
 
// Javascript program for the above approach
 
let mod = 1000000007;
 
// Function to find 2^x using
// modular exponentiation
function exponentMod(A, B)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
 
    // If B is even
    let y;
    if (B % 2 == 0) {
        y = exponentMod(A, B / 2);
        y = (y * y) % mod;
    }
 
    // If B is odd
    else {
        y = A % mod;
        y = (y * exponentMod(A, B - 1)
            % mod)
            % mod;
    }
 
    return ((y + mod) % mod);
}
 
// Function to count the number of
// ways to form the group having
// one leader
function countWays(N)
{
 
    // Find 2^(N-1) using modular
    // exponentiation
    let select = exponentMod(2,
                                N - 1);
 
    // Count total ways
    let ways
        = ((N % mod)
        * (select % mod));
 
    ways %= mod;
 
    // Print the total ways
    document.write(ways);
}
 
// Driver Code
 
    // Given N number of peoples
    let N = 5;
 
    // Function Call
    countWays(N);
 
// This code is contributed by Mayank Tyagi
 
</script>

Output:

Time Complexity: O(log N)
Auxiliary Space: O(N)

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