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

Consider a coding competition on geeksforgeeks practice. Now there 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

**Time Complexity :** O( x )

**Space Complexity :** O( x )

This article is contributed by **nikunj_agarwal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.