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

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 : ^{odd}C_{2} * even.

Case 2 – No. of ways will be : ^{even}C_{3}.

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.` `?>` |

## Javascript

`<script>` `// Javascript program for above implementation` `let 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

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