Given a number N, the task is to find the number of pairs containing an even and an odd number from numbers between 1 and N inclusive. **Note:** The order of numbers in the pair does not matter. That is (1, 2) and (2, 1) are the same.

**Examples**:

Input: N = 3 Output: 2 The pairs are (1, 2) and (2, 3). Input: N = 6 Output: 9 The pairs are (1, 2), (1, 4), (1, 6), (2, 3), (2, 5), (3, 4), (3, 6), (4, 5), (5, 6).

**Approach:** The number of ways to form the pairs is **(Total number of Even numbers*Total number of Odd numbers)**.

Thus

- if N is an even number of even numbers = number of odd numbers = N/2
- if N is an odd number of even numbers = N/2 and the number of odd numbers = N/2+1

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach` `#include <iostream>` `using` `namespace` `std;` `// Driver code` `int` `main()` `{` ` ` `int` `N = 6;` ` ` `int` `Even = N / 2 ;` ` ` `int` `Odd = N - Even ;` ` ` ` ` `cout << Even * Odd ;` ` ` ` ` `return` `0;` ` ` `// This code is contributed` ` ` `// by ANKITRAI1` `}` |

## Java

`// Java implementation of the above approach` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;` `class` `GFG{` `// Driver code` `public` `static` `void` `main(String args[])` `{` ` ` `int` `N = ` `6` `;` ` ` ` ` `int` `Even = N / ` `2` `;` ` ` ` ` `int` `Odd = N - Even ;` ` ` ` ` `System.out.println( Even * Odd );` ` ` `}` `}` |

## Python3

`# Python implementation of the above approach` `N ` `=` `6` ` ` `# number of even numbers` `Even ` `=` `N` `/` `/` `2` `# number of odd numbers` `Odd ` `=` `N` `-` `Even` `print` `(Even ` `*` `Odd)` |

## C#

`// C# implementation of the` `// above approach` `using` `System;` `class` `GFG` `{` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `N = 6;` ` ` ` ` `int` `Even = N / 2 ;` ` ` ` ` `int` `Odd = N - Even ;` ` ` ` ` `Console.WriteLine(Even * Odd);` `}` `}` `// This code is contributed` `// by Akanksha Rai(Abby_akku)` |

## PHP

`<?php` `// PHP implementation of the` `// above approach` `// Driver code` `$N` `= 6;` `$Even` `= ` `$N` `/ 2 ;` `$Odd` `= ` `$N` `- ` `$Even` `;` ` ` `echo` `$Even` `* ` `$Odd` `;` ` ` `// This code is contributed` `// by ChitraNayal` `?>` |

## Javascript

`<script>` `// Javascript implementation of the above approach ` ` ` ` ` `// Driver code` ` ` `let N = 6;` ` ` ` ` `let Even = Math.floor(N / 2) ;` ` ` ` ` `let Odd = N - Even ;` ` ` ` ` `document.write( Even * Odd );` ` ` `// This code is contributed by avanitrachhadiya2155` `</script>` |

**Output:**

9

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