Visualize!

Managing information is hard! There are many problems which concern a number of certain things that match certain criteria, which can simply be a mess to solve algebraically. Luckily, Venn diagrams allow us to organize our counting and effectively solve such types of problems.

Sets

Throughout this module, you should know a little bit of set notation. To give a refresher:

A set is a collection of distinct objects, called its elements, written inside curly braces like $A = \{1, 2, 3\}$. Order doesn't matter and repeats don't count.

1. $x \in A$ means "$x$ is an element of $A$"; $x \notin A$ means it isn't.
2. $|A|$ is the cardinality — the number of elements in $A$.
3. $\varnothing$ is the empty set, with $|\varnothing| = 0$.
4. $A \subseteq B$ means every element of $A$ is also in $B$ ($A$ is a subset of $B$).
5. $U$ is the universal set: everything under consideration in the problem.
6. $A \cup B$ (union) is everything in $A$ or $B$; $A \cap B$ (intersection) is everything in both.
7. $\overline{A}$ (complement) is everything in $U$ not in $A$, so $|\overline{A}| = |U| - |A|$.
8. A two-circle Venn diagram has four regions: only $A$, only $B$, both $A$ and $B$, and neither.
9. $|A \cup B| = |A| + |B| - |A \cap B|$ (inclusion-exclusion for two sets).

Use Cases

In roughly 90% of problems involving several types of criteria (people who take a certain amount of subjects, a deck of cards, etc.), Venn diagrams help in effectively and efficiently solving the problem. However, there is the occasional problem that can be solved quite intuitively without needing a Venn diagram to visualize.

The key steps you should take are:

1. Initialize categories as circles (usually named with letters: A, B, C, ...)
2. Overlap the categories (sometimes two categories shouldn't be overlapped if the problem explicitly states that; however, in your diagram it is still fine to overlap them as long as you remember that the overlap may have length 0)
3. Use conditions given in the problem to label areas (areas being regions in the diagram where a certain number of categories have that property, i.e., a region of overlap)

Techniques

Here are a couple of useful techniques that are nice to know for pattern recognition:

Inclusion-Exclusion

For two sets,

and for three,

The pattern continues for more sets: add all single sets, subtract all pairwise intersections, add all triples, subtract all quadruples, and so on, alternating signs. This is also known as the Principle of Inclusion-Exclusion or PIE.

Complementary Counting

Often "at least one" is annoying to count directly, but "none" is easy. Since every element is either in $A \cup B$ or outside it,

Whenever a problem gives you the "neither" count (or asks for it), use this!

Watch the Wording

A few phrases that trip people up:

• "$|A|$" usually means everyone in $A$, including those also in $B$ — not the "only $A$" region.
• "exactly one of $A$, $B$" means $|A| + |B| - 2|A \cap B|$, since the overlap is removed from both $|A|$ and $|B|$ — that is, twice.
• "at least two of $A$, $B$, $C$" means add the three pairwise overlaps but subtract $2|A \cap B \cap C|$, since the triple region was counted three times.

Demonstration

Like most concepts, Venn diagrams are best explained through example problems. Consider this one:

(2002 AMC 8) At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results: $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues?

This is the inclusion-exclusion formula run backwards: we are handed the two totals and the "neither" count, and asked for the overlap.

First, the number of students who voted in favor of at least one issue is everyone except the $29$ who opposed both:

Now apply $|A \cup B| = |A| + |B| - |A \cap B|$ and solve for the overlap:

It helps to see where every student lands. Once the overlap is $99$, the "only Issue 1" region is $149 - 99 = 50$ and the "only Issue 2" region is $119 - 99 = 20$. As a check, all four regions sum back to the total: $50 + 99 + 20 + 29 = 198$.

Practice Problems