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

27 Apr, 2021
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.
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++
Java
Python3
C#
PHP
Javascript
// C++ program for above implementation#include <bits/stdc++.h>#define MOD 1000000007using namespace std; // Function to count number of waysint 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 codeint main(){    int n = 10;    cout << countWays(n) << endl;    return 0;}
// java program for above implementationimport 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 code for above implementation MOD = 1000000007 # Function to count number of waysdef 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 coden = 10print(int(countWays(n))) # This code is contributed by "Sharad_Bhardwaj"
// C# program for above implementationusing 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 program for// above implementation $MOD = 1000000007; // Function to count// number of waysfunction countWays($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.?>
<script> // Javascript program for above implementationlet MOD = 1000000007;           // Function to count number of ways    function countWays(N)    {        let 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 count;    }     // Driver code        let n = 10;                 document.write(countWays(n));                 // This code is contributed by code_hunt.</script>
Output:
60
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.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
vt_m
code_hunt
Combinatorial
Combinatorial
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Comments
Old Comments
Count of subsets with sum equal to X
Python program to get all subsets of given size of a set
Print all subsets of given size of a set
Make all combinations of size k
Heap's Algorithm for generating permutations
Count Derangements (Permutation such that no element appears in its original position)
Print all permutations in sorted (lexicographic) order
Find all distinct subsets of a given set using BitMasking Approach
Print all distinct permutations of a given string with duplicates
Combinations with repetitions