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

n(Exactly 1 channel) = 175, so your answer is C.

As you can see, fluency with the triple-set venn diagram formulas can be an important skill for GMAT quantitative mastery. It is also critically important on 3-set venn problems to read carefully, as the difference between “at least two” and “exactly two,” for example, is a small change in diction but a massive change in the correct answer. This of course is also true on 2-set venn diagram problems, although the 3-set venn variety just offers even more permutations on how information can be provided and questions can be asked.

Importantly, make sure you know the most standard 3-set venn diagram formulas (the first two on this list) as they are the foundation for all the other 3-set formulas and they’re very difficult and time-consuming to try to “teach yourself” on test day. It’s also extremely helpful to be familiar with the other 3-set formulas from this post so that even if you don’t have them memorized, you’ve seen them and can derive them as necessary from the must-know 3-set venn formulas at the top of the list.

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*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America (wait, could that in and of itself be a 3-set venn diagram problem setup?). She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*