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 ā¦, 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 āPRACTICEā 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 even = 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. ?> |
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
Time Complexity: O(1)
Auxiliary Space: O(1)
This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@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.
Please Login to comment...