Consider a coding competition on geeksforgeeks practice. Now their are **n** distinct participants taking part in the competition. A single participant can make pair with at most one other participant. We need count the number of ways in which **n** participants participating in the coding competition.

**Examples :**

Input : n = 2 Output : 2 2 shows that either both participant can pair themselves in one way or both of them can remain single. Input : n = 3 Output : 4 One way : Three participants remain single Three More Ways : [(1, 2)(3)], [(1), (2,3)] and [(1,3)(2)]

1) Every participant can either pair with another participant or can remain single.

2) Let us consider **X-th** participant, he can either remain single or

he can pair up with someone from **[1, x-1]**.

## C++

// Number of ways in which participant can take part. #include<iostream> using namespace std; int numberOfWays(int x) { // Base condition if (x==0 || x==1) return 1; // A participant can choose to consider // (1) Remains single. Number of people // reduce to (x-1) // (2) Pairs with one of the (x-1) others. // For every pairing, number of people // reduce to (x-2). else return numberOfWays(x-1) + (x-1)*numberOfWays(x-2); } // Driver code int main() { int x = 3; cout << numberOfWays(x) << endl; return 0; }

## Java

// Number of ways in which // participant can take part. import java.io.*; class GFG { static int numberOfWays(int x) { // Base condition if (x==0 || x==1) return 1; // A participant can choose to consider // (1) Remains single. Number of people // reduce to (x-1) // (2) Pairs with one of the (x-1) others. // For every pairing, number of people // reduce to (x-2). else return numberOfWays(x-1) + (x-1)*numberOfWays(x-2); } // Driver code public static void main (String[] args) { int x = 3; System.out.println( numberOfWays(x)); } } // This code is contributed by vt_m.

## Python3

# Python program to find Number of ways # in which participant can take part. # Function to calculate number of ways. def numberOfWays (x): # Base condition if x == 0 or x == 1: return 1 # A participant can choose to consider # (1) Remains single. Number of people # reduce to (x-1) # (2) Pairs with one of the (x-1) others. # For every pairing, number of people # reduce to (x-2). else: return (numberOfWays(x-1) + (x-1) * numberOfWays(x-2)) # Driver code x = 3 print (numberOfWays(x)) # This code is contributed by "Sharad_Bhardwaj"

## C#

// Number of ways in which // participant can take part. using System; class GFG { static int numberOfWays(int x) { // Base condition if (x == 0 || x == 1) return 1; // A participant can choose to // consider (1) Remains single. // Number of people reduce to // (x-1) (2) Pairs with one of // the (x-1) others. For every // pairing, number of people // reduce to (x-2). else return numberOfWays(x - 1) + (x - 1) * numberOfWays(x - 2); } // Driver code public static void Main () { int x = 3; Console.WriteLine(numberOfWays(x)); } } // This code is contributed by vt_m.

## PHP

<?php // Number of ways in which // participant can take part. function numberOfWays($x) { // Base condition if ($x == 0 || $x == 1) return 1; // A participant can choose // to consider (1) Remains single. // Number of people reduce to (x-1) // (2) Pairs with one of the (x-1) // others. For every pairing, number // of peopl reduce to (x-2). else return numberOfWays($x - 1) + ($x - 1) * numberOfWays($x - 2); } // Driver code $x = 3; echo numberOfWays($x); // This code is contributed by Sam007 ?>

**Output :**

4

Since there are overlapping subproblems, we can optimize it using dynamic programming.

## C++

// Number of ways in which participant can take part. #include<iostream> using namespace std; int numberOfWays(int x) { int dp[x+1]; dp[0] = dp[1] = 1; for (int i=2; i<=x; i++) dp[i] = dp[i-1] + (i-1)*dp[i-2]; return dp[x]; } // Driver code int main() { int x = 3; cout << numberOfWays(x) << endl; return 0; }

## Java

// Number of ways in which // participant can take part. import java.io.*; class GFG { static int numberOfWays(int x) { int dp[] = new int[x+1]; dp[0] = dp[1] = 1; for (int i=2; i<=x; i++) dp[i] = dp[i-1] + (i-1)*dp[i-2]; return dp[x]; } // Driver code public static void main (String[] args) { int x = 3; System.out.println(numberOfWays(x)); } } // This code is contributed by vt_m.

## Python3

# Python program to find Number of ways # in which participant can take part. # Function to calculate number of ways. def numberOfWays (x): # Base condition if x == 0 or x == 1: return 1 # A participant can choose to consider # (1) Remains single. Number of people # reduce to (x-1) # (2) Pairs with one of the (x-1) others. # For every pairing, number of people # reduce to (x-2). else: return (numberOfWays(x-1) + (x-1) * numberOfWays(x-2)) # Driver code x = 3 print (numberOfWays(x)) # This code is contributed by "Sharad_Bhardwaj"

## C#

// Number of ways in which // participant can take part. using System; class GFG { static int numberOfWays(int x) { int []dp = new int[x+1]; dp[0] = dp[1] = 1; for (int i = 2; i <= x; i++) dp[i] = dp[i - 1] + (i - 1) * dp[i - 2]; return dp[x]; } // Driver code public static void Main () { int x = 3; Console.WriteLine(numberOfWays(x)); } } // This code is contributed by vt_m.

## PHP

<?php // PHP program for Number of ways // in which participant can take part. function numberOfWays($x) { $dp[0] = 1; $dp[1] = 1; for ($i = 2; $i <= $x; $i++) $dp[$i] = $dp[$i - 1] + ($i - 1) * $dp[$i - 2]; return $dp[$x]; } // Driver code $x = 3; echo numberOfWays($x) ; // This code is contributed by Sam007 ?>

Output:

4

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