What is the time complexity of the following code that determines the number of inversions in an array?

int InvCount(int arr[], int n) 
{ 
	int count = 0; 
	for (int i = 0; i < n - 1; i++) 
		for (int j = i + 1; j < n; j++) 
			if (arr[i] > arr[j]) 
				count++; 
 
	return count; 
}

A. O(n)

B. O(n log n)

C. O(n²)

D. O(log n)

Answer: Option C