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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 <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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