Number of ways to pair people

Given that there are p people in a party. Each person can either join dance as a single individual or as a pair with any other. The task is to find the number of different ways in which p people can join the dance.
Examples:

Input : p = 3
Output : 4
Let the three people be P1, P2 and P3
Different ways are: {P1, P2, P3}, {{P1, P2}, P3},
{{P1, P3}, P2} and {{P2, P3}, P1}.

Input : p = 2
Output : 2
The groups are: {P1, P2} and {{P1, P2}}.

Approach: The idea is to use dynamic programming to solve this problem. There are two situations: Either the person join dance as single individual or as a pair. For the first case the problem reduces to finding the solution for p-1 people. For the second case, there are p-1 choices to select an individual for pairing and after selecting an individual for pairing the problem reduces to finding solution for p-2 people as two people among p are already paired. 
So the formula for dp is: 

dp[p] = dp[p-1] + (p-1) * dp[p-2].

Below is the implementation of the above approach: 

C++

// CPP program to find number of ways to
// pair people in party

#include <bits/stdc++.h>
using namespace std;

// Function to find number of ways to
// pair people in party
int findWaysToPair(int p)
{
    // To store count of number of ways.
    int dp[p + 1];

    dp[1] = 1;
    dp[2] = 2;

    // Using the recurrence defined find
    // count for different values of p.
    for (int i = 3; i <= p; i++) {
        dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
    }

    return dp[p];
}

// Driver code
int main()
{
    int p = 3;
    cout << findWaysToPair(p);
    return 0;
}

Java

// Java program to find number of ways to
// pair people in party

class GFG
{

// Function to find number of ways to
// pair people in party
static int findWaysToPair(int p)
{
    // To store count of number of ways.
    int dp[] = new int[p + 1];

    dp[1] = 1;
    dp[2] = 2;

    // Using the recurrence defined find
    // count for different values of p.
    for (int i = 3; i <= p; i++)
    {
        dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
    }

    return dp[p];
}

// Driver code
public static void main(String args[])
{
    int p = 3;
    System.out.println(findWaysToPair(p));
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to find number of
# ways to pair people in party

# Function to find number of ways
# to pair people in party
def findWays(p):

    # To store count of number of ways.
    dp = [0] * (p + 1)
    dp[1] = 1
    dp[2] = 2

    # Using the recurrence defined find
    # count for different values of p.
    for i in range(3, p + 1):
        dp[i] = (dp[i - 1] +
                   (i - 1) * dp[i - 2])
    return dp[p]

# Driver code
p = 3
print(findWays(p))

# This code is contributed by Shrikant13

C#

// C# program to find number of ways to
// pair people in party
using System;

class GFG
{

// Function to find number of ways to
// pair people in party
public static int findWaysToPair(int p)
{
    // To store count of number of ways.
    int[] dp = new int[p + 1];

    dp[1] = 1;
    dp[2] = 2;

    // Using the recurrence defined find
    // count for different values of p.
    for (int i = 3; i <= p; i++)
    {
        dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
    }
    return dp[p];
}

// Driver code
public static void Main(string[] args)
{
    int p = 3;
    Console.WriteLine(findWaysToPair(p));
}
}

// This code is contributed by shrikanth13

PHP

<?php
// PHP program to find number of ways
// to pair people in party

// Function to find number of ways to
// pair people in party
function findWaysToPair($p)
{
    // To store count of number of ways.
    $dp = array();

    $dp[1] = 1;
    $dp[2] = 2;

    // Using the recurrence defined find
    // count for different values of p.
    for ($i = 3; $i <= $p; $i++)
    {
        $dp[$i] = $dp[$i - 1] +
                     ($i - 1) * $dp[$i - 2];
    }

    return $dp[$p];
}

// Driver code
$p = 3;
echo findWaysToPair($p);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>
// Javascript program to find number of ways to
// pair people in party

// Function to find number of ways to
// pair people in party
function findWaysToPair(p)
{
    // To store count of number of ways.
    var dp = Array(p+1);

    dp[1] = 1;
    dp[2] = 2;

    // Using the recurrence defined find
    // count for different values of p.
    for (var i = 3; i <= p; i++) {
        dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
    }

    return dp[p];
}

// Driver code
var p = 3;
document.write( findWaysToPair(p));

// This code is contributed by noob2000.
</script>

Output:

4

Time Complexity: O(p) 
Auxiliary Space: O(p)