# Minimum value possible of a given function from the given set

Given a set Sn that represents a finite set {1, 2, 3, ….., n}. Zn denotes the set of all subsets of Sn which contains exactly 2 elements. The task is to find the value of F(n). Examples:

Input: N = 3
Output: 4
For n=3 we get value 1, 2 times and 2, 1 times
thus the answer would be 1 * 2 + 2 * 1 = 4.

Input: N = 10
Output: 165


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

For each value as we go from left to right in the set we get each value, n-value number of times that value.
i.e. For each i, value to be added is (i + 1) * (n – i – 1)

Below is the implementation of above approach:

## C++

 // C++ implementation of above approach  #include  using namespace std;  #define ll long long     // Function to find the value of F(n)  ll findF_N(ll n)  {         ll ans = 0;      for (ll i = 0; i < n; ++i)          ans += (i + 1) * (n - i - 1);         return ans;  }     // Driver code  int main()  {         ll n = 3;      cout << findF_N(n);     return 0;  }

## Java

 // Java implementation of above approach     import java.io.*;     class GFG {               // Function to find the value of F(n)  static long findF_N(long n)  {         long ans = 0;      for (long i = 0; i < n; ++i)          ans += (i + 1) * (n - i - 1);         return ans;  }     // Driver code         public static void main (String[] args) {              long n = 3;      System.out.println( findF_N(n));      }  }  // This code is contributed by anuj_67..

## Python3

 # Python3 implementation of  # above approach     # Function to find the value of F(n)  def findF_N(n):         ans = 0     for i in range(n):          ans = ans + (i + 1) * (n - i - 1)         return ans     # Driver code  n = 3 print(findF_N(n))     # This code is contributed by  # Sanjit_Prasad

## C#

 // C# implementation of above approach  using System;     class GFG  {  // Function to find the   // value of F(n)  static long findF_N(long n)  {         long ans = 0;      for (long i = 0; i < n; ++i)          ans += (i + 1) * (n - i - 1);         return ans;  }     // Driver code  public static void Main ()   {      long n = 3;      Console.WriteLine(findF_N(n));  }  }     // This code is contributed by anuj_67

## PHP

 

Output:

4
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