22.1.2. Extension to Three Sets

So, to find out the cardinality of the union of 3 sets, it is the summation of the cardinalities of the individual sets. Then we have to take the difference of the cardinalities of 2 sets at a time and then again we have to add the cardinality of the intersection of all the 3 sets.

When using the inclusion-exclusion principle for three sets, let's denote these sets as A, B, and C. To compute the total number of unique elements in the union of these sets (A ∪ B ∪ C), we first sum the sizes of each individual set: |A| + |B| + |C|. However, when we do this, we double-count all elements that are common to pairs of sets. Therefore, we need to subtract the sizes of the intersections of each pair: |A ∩ B|, |A ∩ C|, and |B ∩ C|. Finally, if there are elements common to all three sets, we need to add back the size of the intersection of A, B, and C (|A ∩ B ∩ C|) because those elements were subtracted too many times.

Imagine you are organizing three different clubs, and members from each club can also join other clubs. If you simply count the total members from each club, you'd overestimate the total because several members might belong to more than one club. By subtracting the overlapping members, you accurately find how many unique members belong to at least one club.

Because the elements in A ∩ B are counted twice if we add individually the cardinalities of the A set and B set. Similarly, when we are adding up the cardinalities of A set and C set the common portion is included twice and so on.

Each time we include the count from a set, any elements that also belong to other sets get counted again. For example, if a member belongs to both sets A and B, when we add |A| and |B|, we effectively count that member twice. To correct this, we discount those overlapping members by subtracting the sizes of their intersections. This ensures that any element captured in more than one set is only counted once when calculating the total unique count of elements across all sets.

Think of a classroom where some students are enrolled in multiple subjects. If you count the students in Math, English, and Science classes by looking at each list, you'll find some names appear in several lists. They shouldn't be counted multiple times; hence, you need to ensure that their names are only counted once, just like correcting the over-counting with the inclusion-exclusion principle.

The elements which are common to all the 3 sets are counted thrice. If we do not add the cardinality of intersection of A B and C in this overall formula, then the contribution of the common elements in the 3 sets is ignored.

When we only use subtraction to correct the counts for two sets, elements that belong to all three sets may be subtracted too many times. Specifically, if a member belongs to A, B, and C, it gets counted in each individual set's count and may end up being subtracted once for each pair of intersections. Thus, to accurately reflect the total unique members, we must add back the count of those who belong to all three sets, ensuring we do not ignore their contributions.

Consider a buffet where some dishes are available in multiple categories, like a dessert that counts as both a fruit and a sweet. If you count only the dishes within each category separately, some of the items might seem to disappear when you try to balance your dish count. By acknowledging that you have to add back those multi-category dishes, you ensure you account for them just once in your overall meal list.

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