Given an integer array, count the total number of triplets, which leads to an inversion. If (i < j < k) and (nums[i] > nums[j] > nums[k]), then the triplet (i, j, k) has formed an inversion in array `nums`.

int nums[] = {9, 6, 4, 5, 8};

int count = 0;
for (int i = 0; i < nums.length; i++) {
  for (int j = i + 1; j < nums.length; j++) {
    for (int k = j + 1; k < nums.length; k++) {
      if (nums[i] > nums[j] && nums[j] > nums[k]) {
        count++;
      }
    }
  }
}
return count;