# Inclusion and Exclusion and the new GMAT

Dublin native Colm Mulcahy has been in the Department of Mathematics at Spelman College since 1988. His interests include algebra, number theory, geometry and mathematical card principles and effects. Follow him on Twitter at @CardColm and also check out @WWMGT

The last question, under the heading “Two-Part Analysis”, at the end of this NYT article (from July 2011) on the new GMAT seems to be deliberately worded in a way that forces one to read and think very carefully.

It takes a while to even process the question as it’s asked! I’m assuming that was intentional.

I’m curious how “they” intended people to solve this. Exclude impossible answers until only one is still Included?  I guess so.

Yes, it can also be solved quickly using the Inclusion Exclusion principle (click for a spoiler).

Spoiler:  Inclusion Exclusion Principle says $|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |B \cap C| + |A \cap B \cap C|.$   Since there are 20 people in each of $A$, $B$, $C$ individually, the only possible number given in the table for the total number of people is $34$.   But now we have $34 = 20 + 20 + 20 - 10 - 10 - 10 + |A \cap B \cap C|$ which gives $|A \cap B \cap C| = 4.$

My mathematician friend Peter Ross from Santa Clara College in California comments:

Inclusion-exclusion isn’t needed, given that the answers must be among the 5 given. In fact, using two 3-set Venn Diagrams, one with 4 in the triple intersection and one with 8 in it (the only possible values since the double intersections have only 10), just fill out uniquely the other regions. The first case yields the answers 4 and 34 total, while the second case gives 8 and 38 total, so it’s excluded and the answers are 4 and 34. I imagine it’s called a two-part problem as there are two cases.  Moral–think visually not symbolically!

Algebra can play a role too! The 3 intersecting sets Venn diagram has 8 regions, one of which [the complement of the triple union] we are told has no elements in it. Filling in the rest using $t$ for the count in the triple intersection, $d$ for the (necessarily identical) double (only) intersection counts and $s$ for the (necessarily identical) counts of the elements in just one single set, we get $t + d = 10$ for each double intersection, and $t + 2d + s = 20$ for each individual set.

Subtracting, we see that $d + s = 10$ also, and comparing with the above we deduce that $t = s$.

Now use the fact of the choices on offer in the table, the total number of people (which is at least 20) is now $4t + 3d$ and must equal 34. Since $t + d = 10$ we also have $3t + 3d = 30$, and subtracting yields $t = 4$ as desired.

Thanks to my Econ colleague Jack Stone at Spelman for passing this along.

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