# 7 Formulas for Tackling Three Overlapping Sets on the GMAT Venn diagram problems are a staple of the GMAT word problems repertoire, and while people are generally familiar with the 2-set venn diagram formulas, the 3-set venn diagram equations often pose a challenge for test-takers. In a previous post, we saw how to solve three overlapping sets questions using venn diagrams. Today, we will look at all of the various venn diagram formulas floating around on three overlapping sets. Most of these are self explanatory but we will look into the details of the less commonly known three-set venn formulas.

There are two basic 3-set venn diagram formulas that we already know:

1) Total = n(No Set) + n(Exactly one set) + n(Exactly two sets) + n(Exactly three sets)

2) Total = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) + n(No Set)

From these two standard venn formulas, we can derive all the other three-set venn formula offshoots:

n(Exactly one set) + n(Exactly two sets) + n(Exactly three sets) gives us n(At least one set). So we get:

3) Total = n(No Set) + n(At least one set)

From (3), we get n(At least one set) = Total – n(No Set)

Plugging this into (2), we then get:

4) n(At least one set) = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C)

Now let’s see how we can calculate the number of people in exactly two sets. There is a reason we jumped to n(Exactly two sets) instead of following the more logical next step of figuring out n(At least two sets) – it will be more intuitive to get n(At least two sets) after we find n(Exactly two sets).

n(A and B) includes people who are in both A and B and it also includes people who are in A, B and C. Because of this, we should remove n(A and B and C) from n(A and B) to get n(A and B only). Similarly, you get n(B and C only) and n(C and A only), so adding all these three will give us number of people in exactly 2 sets.

n(Exactly two sets) = n(A and B) – n(A and B and C) + n(B and C) – n(A and B and C) + n(C and A) – n(A and B and C). Therefore:

5) n(Exactly two sets) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C)

Now we can easily get n(At least two sets):

6) n(At least two sets) = n(A and B) + n(B and C) + n(C and A) – 2*n(A and B and C)

This is just n(A and B and C) more than n(Exactly two sets). That makes sense, doesn’t it? Here, you include the people who are in all three sets once and n(Exactly two sets) converts to n(At least two sets)!

Now, we go on to find n(Exactly one set). From n(At least one set), let’s subtract n(At least two sets); i.e. we subtract (6) from (4)

n(Exactly one set) = n(At least one set) – n(At least two sets), therefore:

7) n(Exactly one set) = n(A) + n(B) + n(C) – 2*n(A and B) – 2*n(B and C) – 2*n(C and A) + 3*n(A and B and C)

You don’t need to learn all these formulas. Just focus on first two and know how you can arrive at the others if required. Let’s try this in an example problem:

Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

(A) 185

(B) 180

(C) 175

(D) 190

(E) 195

You are given that:

n(At least one channel) = 250

n(Exactly two channels) = 50

So we know that n(At least one channel) = n(Exactly 1 channel) + n(Exactly 2 channels) + n(Exactly 3 channels) = 250

250 = n(Exactly 1 channel) + 50 + n(Exactly 3 channels)

Let’s find the value of n(Exactly 3 channels) = x

We also know that n(At least one channel) = n(A) + n(B) + n(C) – n(A and B) – n(B and C) – n(C and A) + n(A and B and C) = 250

Also, n(Exactly two channels) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C)

So n(A and B) + n(B and C) + n(C and A) = n(Exactly two channels) + 3*n(A and B and C)

Plugging this into the equation above:

250 = n(A) + n(B) + n(C) – n(Exactly two channels) – 3*x + x

250 = 116 + 127 + 107 – 50 – 2x

x = 25

250 = n(Exactly 1 channel) + 50 + 25