Hello, coding enthusiast! In our journey to master coding and problem-solving, we've arrived at an interesting challenge today. We're going to focus heavily on combinatorial problems in practice. Specifically, we're examining combinatorial problems that involve working with large data sets and multiple pairs of numbers. We'll learn how to solve significant problems efficiently by implementing smart use of data structures like hashes and sidestepping expensive operations like iteration over large arrays. Are you ready? Let's dive in!

In this unit's task, you'll be given a large array composed of pairs of distinct, positive integers, including up to 1,000,000 elements. Your challenge is to write a Ruby method to count the number of indices (i, j) (i != j) where the i-th pair does not share a common element with the j-th pair. A crucial point to remember is that a pair (a, b) is considered identical to (b, a), meaning the order of elements in a pair is irrelevant in this case. It is guaranteed that no two pairs are element-wise equal.

For example, given the array [[2, 5], [1, 6], [3, 2], [4, 2], [5, 1], [6, 3]], the output should be 8. The required index pairs are the following: (0, 1) (i.e., the pair does not share a common element with pair ), ( does not share a common element with ), , , , , , .