# Number of ways to choose a pair containing an even and an odd number from 1 to N

• Last Updated : 20 May, 2021

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

1. if N is an even number of even numbers = number of odd numbers = N/2
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 using namespace std; // Driver codeint 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 approachimport java.util.*;import java.lang.*;import java.io.*;class GFG{ // Driver codepublic 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 approachN = 6  # number of even numbersEven = N//2 # number of odd numbersOdd = N-Evenprint(Even * Odd)

## C#

 // C# implementation of the// above approachusing System; class GFG{ // Driver codepublic 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)



## Javascript



Output:

9

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