Given an integer N, find the number of ways we can choose 3 numbers from {1, 2, 3 …, N} such that their sum is even.
Examples:

Input :  N = 3
Output : 1
Explanation: Select 1, 2 and 3

Input :  N = 4
Output :  2
Either select (1, 2, 3) or (1, 3, 4)

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

To get sum even there can be only 2 cases:

Take 2 odd numbers and 1 even.
Take all even numbers.

If n is even,
  Count of odd numbers = n/2 and even = n/2.
Else
  Count odd numbers = n/2 +1 and evne = n/2.

Case 1 – No. of ways will be : ^oddC₂ * even.
Case 2 – No. of ways will be : ^evenC₃.

So, total ways will be Case_1_result + Case_2_result.

C++

// C++ program for above implementation 
#include <bits/stdc++.h> 
#define MOD 1000000007 
using namespace std; 
  
// Function to count number of ways 
int countWays(int N) 
{ 
    long long int count, odd = N / 2, even; 
    if (N & 1) 
        odd = N / 2 + 1; 
  
    even = N / 2; 
  
    // Case 1: 2 odds and 1 even 
    count = (((odd * (odd - 1)) / 2) * even) % MOD; 
  
    // Case 2: 3 evens 
    count = (count + ((even * (even - 1) *  
                           (even - 2)) / 6)) % MOD; 
  
    return count; 
} 
  
// Driver code 
int main() 
{ 
    int n = 10; 
    cout << countWays(n) << endl; 
    return 0; 
} 

Java

// java program for above implementation 
import java.io.*; 
  
class GFG { 
      
    static long MOD = 1000000007; 
      
    // Function to count number of ways 
    static long countWays(int N) 
    { 
        long count, odd = N / 2, even; 
          
        if ((N & 1) > 0) 
            odd = N / 2 + 1; 
      
        even = N / 2; 
      
        // Case 1: 2 odds and 1 even 
        count = (((odd * (odd - 1)) / 2) 
                          * even) % MOD; 
      
        // Case 2: 3 evens 
        count = (count + ((even * (even 
                - 1) * (even - 2)) / 6)) 
                                  % MOD; 
      
        return (long)count; 
    } 
      
    // Driver code 
    static public void main (String[] args) 
    { 
        int n = 10; 
          
        System.out.println(countWays(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

Python3

# Python3 code for above implementation 
  
MOD = 1000000007
  
# Function to count number of ways 
def countWays( N ): 
    odd = N / 2
    if N & 1: 
        odd = N / 2 + 1
    even = N / 2
      
    # Case 1: 2 odds and 1 even 
    count = (((odd * (odd - 1)) / 2) * even) % MOD 
  
    # Case 2: 3 evens 
    count = (count + ((even * (even - 1) *
            (even - 2)) / 6)) % MOD 
    return count 
  
# Driver code 
n = 10
print(int(countWays(n))) 
  
# This code is contributed by "Sharad_Bhardwaj" 

C#

// C# program for above implementation 
using System; 
  
public class GFG { 
      
    static long MOD = 1000000007; 
      
    // Function to count number of ways 
    static long countWays(int N) 
    { 
        long count, odd = N / 2, even; 
          
        if ((N & 1) > 0) 
            odd = N / 2 + 1; 
      
        even = N / 2; 
      
        // Case 1: 2 odds and 1 even 
        count = (((odd * (odd - 1)) / 2)  
                            * even) % MOD; 
      
        // Case 2: 3 evens 
        count = (count + ((even * (even  
                  - 1) * (even - 2)) / 6)) 
                                    % MOD; 
      
        return (long)count; 
    } 
      
    // Driver code 
    static public void Main () 
    { 
        int n = 10; 
  
        Console.WriteLine(countWays(n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program for  
// above implementation 
  
$MOD = 1000000007; 
  
// Function to count  
// number of ways 
function countWays($N) 
{ 
    global $MOD; 
      
    $count;  
    $odd =$N / 2;  
    $even; 
    if ($N & 1) 
        $odd = $N / 2 + 1; 
  
    $even = $N / 2; 
  
    // Case 1: 2 odds  
    // and 1 even 
    $count = ((($odd * ($odd - 1)) / 2) *  
                            $even) % $MOD; 
  
    // Case 2: 3 evens 
    $count = ($count + (($even * ($even - 1) *  
                        ($even - 2)) / 6)) % $MOD; 
  
    return $count; 
} 
  
    // Driver Code 
    $n = 10; 
    echo countWays($n); 
  
// This code is contributed by anuj_67. 
?> 

Output:

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

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My Personal Notes

Number of ways to get even sum by choosing three numbers from 1 to N

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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