In this triple overlapping question, a shortcut is presented to subtract the duplicates. However, I don't understand the logic they used to subtract the duplicates from the the 3 clubs. I've highlighted this part of the solution in RED.
Here's the question: Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?
Here's the solution: If we add up the total number of club sign-ups, or registrations, we get 22 + 27 + 28 = 77. We must remember that this number includes overlapping registrations (some students sign up for two clubs, others for three). So, there are 77 registrations and 59 total students. Therefore, there must be 77 - 59 = 18 duplicate registrations.
We know that 6 of these duplicates come from those 6 students who sign up for exactly two clubs. Each of these 6, then, adds one extra registration, for a total of 6 duplicates. We are then left with 18 - 6 = 12 duplicate registrations. These 12 duplicates must come from those students who sign up for all three clubs.
For each student who signs up for three clubs, there are two extra sign-ups. Therefore, there must be 6 students who sign up for three clubs:
12 duplicates / (2 duplicates/student) = 6 students
Between the 6 students who sign up for two clubs and the 6 students who sign up for all three, we have accounted for all 18 duplicate registrations.
So, the number of students who sign up for all three clubs is 6.