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Count of ways to form 2 necklace from N beads containing N/2 beads each

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  • Last Updated : 07 Oct, 2022
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Given a positive even integer N denoting the number of distinct beads, the task is to find the number of ways to make 2 necklaces having exactly N/2 beads.

Examples:

Input: N = 2
Output: 1
Explanation: 
The only possible way to make two necklaces is that {1 | 2}.

Input: N = 4
Output: 3
Explanation: 
The possible ways to make two necklaces are {(1, 2) | (3, 4)}, {(1, 3) | (2, 4)}, and {(1, 4) | (2, 3)}.

Approach: The problem can be solved by using the concept of circular permutation and combinatorics. Follow the steps below to solve the problem:

  • Define a function, say factorial for calculating the factorial of a number, by following the steps below:
    • Base Case: If n = 0, then return 1.
    • If n != 0, then recursively call the function and return n * factorial(n-1).
  • Initialize a variable, say, ans as C(N, N/2), that is the number of ways to choose N/2 beads from N beads.
  • Since the necklace is circular, the number of ways to permute N/2 beads are factorial(N/2 -1), so multiply the value of ans by factorial(N/2 -1)*factorial(N/2-1) since there are two necklaces.
  • Now divide the ans by 2. Because of symmetrical distribution. For eg, For N=2, the distribution {1 | 2} and {2 | 1} are considered the same.
  • Finally, after completing the above steps, print the value of ans as the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate factorial
int factorial(int n)
{
    if (n == 0)
        return 1;
    return n * factorial(n - 1);
}
 
// Function to count number of ways
// to make 2 necklace having exactly
// N/2 beads if each bead is
// considered different
long long numOfNecklace(int N)
{
    // Number of ways to choose N/2 beads
    // from N beads
    long long ans = factorial(N)
                    / (factorial(N / 2) * factorial(N / 2));
 
    // Number of ways to permute N/2 beads
    ans = ans * factorial(N / 2 - 1);
    ans = ans * factorial(N / 2 - 1);
 
    // Divide ans by 2 to remove repetitions
    ans /= 2;
 
    // Return ans
    return ans;
}
 
// Driver Code
int main()
{
    // Given Input
    int N = 4;
 
    // Function Call
    cout << numOfNecklace(N) << endl;
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
class GFG
{
   
    // Function to calculate factorial
    static int factorial(int n)
    {
        if (n == 0)
            return 1;
        return n * factorial(n - 1);
    }
   
    // Function to count number of ways
    // to make 2 necklace having exactly
    // N/2 beads if each bead is
    // considered different
    static long numOfNecklace(int N)
    {
       
        // Number of ways to choose N/2 beads
        // from N beads
        long ans = factorial(N)
                   / (factorial(N / 2) * factorial(N / 2));
 
        // Number of ways to permute N/2 beads
        ans = ans * factorial(N / 2 - 1);
        ans = ans * factorial(N / 2 - 1);
 
        // Divide ans by 2 to remove repetitions
        ans /= 2;
 
        // Return ans
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
       
        // Given Input
        int N = 4;
 
        // Function Call
        System.out.println(numOfNecklace(N));
 
    }
}
 
// This code is contributed by Potta Lokesh

Python3




# Python 3 program for the above approach
 
# Function to calculate factorial
def factorial(n):
    if (n == 0):
        return 1
    return n * factorial(n - 1)
 
# Function to count number of ways
# to make 2 necklace having exactly
# N/2 beads if each bead is
# considered different
def numOfNecklace(N):
   
    # Number of ways to choose N/2 beads
    # from N beads
    ans = factorial(N) // (factorial(N // 2) * factorial(N // 2))
 
    # Number of ways to permute N/2 beads
    ans = ans * factorial(N // 2 - 1)
    ans = ans * factorial(N // 2 - 1)
 
    # Divide ans by 2 to remove repetitions
    ans //= 2
 
    # Return ans
    return ans
 
# Driver Code
if __name__ == '__main__':
    # Given Input
    N = 4
 
    # Function Call
    print(numOfNecklace(N))
     
    # This code is contributed by ipg2016107.

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to calculate factorial
static int factorial(int n)
{
    if (n == 0)
        return 1;
         
    return n * factorial(n - 1);
}
 
// Function to count number of ways
// to make 2 necklace having exactly
// N/2 beads if each bead is
// considered different
static long numOfNecklace(int N)
{
     
    // Number of ways to choose N/2 beads
    // from N beads
    long ans = factorial(N) /
              (factorial(N / 2) *
               factorial(N / 2));
 
    // Number of ways to permute N/2 beads
    ans = ans * factorial(N / 2 - 1);
    ans = ans * factorial(N / 2 - 1);
 
    // Divide ans by 2 to remove repetitions
    ans /= 2;
 
    // Return ans
    return ans;
}
 
// Driver Code
static public void Main ()
{
     
    // Given Input
    int N = 4;
 
    // Function Call
    Console.Write( numOfNecklace(N));
}
}
 
// This code is contributed by sanjoy_62

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to calculate factorial
function factorial(n)
{
    if (n == 0)
        return 1;
    return n * factorial(n - 1);
}
 
// Function to count number of ways
// to make 2 necklace having exactly
// N/2 beads if each bead is
// considered different
function numOfNecklace(N)
{
 
    // Number of ways to choose N/2 beads
    // from N beads
    var ans = factorial(N)
                    / (factorial(N / 2) * factorial(N / 2));
 
    // Number of ways to permute N/2 beads
    ans = ans * factorial(N / 2 - 1);
    ans = ans * factorial(N / 2 - 1);
 
    // Divide ans by 2 to remove repetitions
    ans /= 2;
 
    // Return ans
    return ans;
}
 
// Driver Cod
    // Given Input
    var N = 4;
 
    // Function Call
    document.write(numOfNecklace(N));
 
// This code is contributed by SURENDRA_GANGWAR.
</script>

Output

3

Time Complexity: O(N)
Auxiliary Space: O(N), due to recursive call stack


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