# Counting pairs when a person can form pair with at most one

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` `?>` |

## Javascript

`<script>` `// 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 people` ` ` `// reduce to (x-2).` ` ` `else` ` ` `return` `numberOfWays(x - 1) + (x - 1) * numberOfWays(x - 2);` ` ` `}` ` ` `// Driver code` ` ` `var` `x = 3;` ` ` `document.write(numberOfWays(x));` `// This code is contributed by gauravrajput1 ` `</script>` |

**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 vipinyadav15799` |

## Python3

`# Python program to find Number of ways ` `# in which participant can take part.` `# Function to calculate number of ways.` `def` `numberOfWays (x):` ` ` `dp` `=` `[]` ` ` `dp.append(` `1` `)` ` ` `dp.append(` `1` `)` ` ` `for` `i ` `in` `range` `(` `2` `,x` `+` `1` `):` ` ` `dp.append(dp[i` `-` `1` `]` `+` `(i` `-` `1` `)` `*` `dp[i` `-` `2` `])` ` ` `return` `(dp[x])` ` ` `# 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` `?>` |

## Javascript

`<script>` `// Number of ways in which` `// participant can take part.` ` ` `function` `numberOfWays( x) ` ` ` `{` ` ` `let dp = Array(x + 1).fill(0);` ` ` `dp[0] = dp[1] = 1;` ` ` `for` `( i = 2; i <= x; i++)` ` ` `dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];` ` ` `return` `dp[x];` ` ` `}` ` ` `// Driver code` ` ` `let x = 3;` ` ` `document.write(numberOfWays(x));` `// This code is contributed by gauravrajput1` `</script>` |

Output:

4

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