Three-set counting is often a bookkeeping problem. Before using a formula, establish whether a pairwise overlap includes people who are also in the third set. That definition changes how the data should be interpreted.
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State what the overlap numbers include
In a fictional group of 60 students, 30 belong to club A, 28 to B and 20 to C. The pairwise overlaps are 12 in A and B, 10 in A and C, and 8 in B and C. Each pairwise number includes the five students in all three clubs.
These are invented counts for a teaching example. The sets represent membership, and each student is counted once in the overall group.
Count the union
Adding 30 + 28 + 20 counts students in two clubs twice and those in three clubs three times. Subtract the three pairwise overlaps, then add the triple overlap back once.
The union is 30 + 28 + 20 - 12 - 10 - 8 + 5 = 53. Therefore 60 - 53 = 7 students belong to none of the clubs.
Find membership in A only
Start with all 30 members of A. Subtract the 12 in A and B and the 10 in A and C. The five triple members have now been subtracted twice, so add them back once.
The number in A only is 30 - 12 - 10 + 5 = 13. This answer is different from the total number in exactly one club, which would also require the B-only and C-only regions.
Use the disjoint regions as a check
The pair-only regions are 7 in AB, 5 in AC and 3 in BC. The single-only regions are 13 in A, 13 in B and 7 in C. Together with five in all three, these sum to 53.
Starting with the central overlap and working outward is a reliable diagram method. All region counts should be nonnegative and should reconstruct each given total. This original problem supports AMC 12 preparation; current competition details should be checked through MAA.
Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.
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Official information and further reading: Source 1
