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.

It sounds like you're okay with the part where we get rid of the 6 overcounts from the students who signed up for two clubs, so let's use that as a starting point. This leaves us with 12 overcounts. Well, of course students who sign up for only one club are not overcounted. We've already dealt with the overcounts in the three regions where two clubs overlap, so the remaining overcounts must occur in the area where all three clubs overlap. Each student in that portion of the Venn Diagram has signed up for three clubs and has thus been counted three times, which means 2/3 of the registrations are overcounts and 1/3 are legit. If the 12 overcounts represent 2/3 of the registrations, that gives us 18 registrations in the central region of the Venn Diagram and thus 6 unique students..

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Tim Sanders
Manhattan GMAT Instructor

Dear Instructor,
Is there an easier explanation to this question. The part that takes care of the over lap is confusing and time consuming.
P.S is it OK to conclude since, we are left with 18 over lap, issing number property that the answer must contain at least 2 and 3 in the prime box.
Thanks.

Alexei, I think Tim's explanation is excellent. Go back and read it carefully and try actually drawing this out and representing what is going on. I don't think you can make those conclusions about the numbers in the prime box.

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Jamie Nelson
ManhattanGMAT Instructor