Practical problems on union and intersection

Counting with sets is everywhere , surveys, polls, exam statistics, sports data. The key formula is the inclusion–exclusion principle: if we add the sizes of overlapping sets we overcount the overlap, so we must subtract it back. Drawing a Venn diagram in every problem keeps the bookkeeping clean.

Definitions and formulas

For finite sets $A, B \subseteq U$: $\boxed{|A \cup B| = |A| + |B| - |A \cap B|.}$

For three sets: $\boxed{|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C|.}$

If $A, B$ are disjoint then $|A \cup B| = |A| + |B|$, recovering the easier rule.

Also useful: $|A - B| = |A| - |A \cap B|$ and $|A'| = |U| - |A|$.

Derivation of the two-set formula

Split $A \cup B$ into three disjoint pieces:

• $A - B$ (in $A$ only),
• $B - A$ (in $B$ only),
• $A \cap B$ (in both).

These three are pairwise disjoint and their union is $A \cup B$. So $|A \cup B| = |A - B| + |B - A| + |A \cap B|.$

Now $|A| = |A - B| + |A \cap B|$ (since $A$ splits as "only in $A$" and "in both"). Similarly $|B| = |B - A| + |A \cap B|$. Adding, $|A| + |B| = |A - B| + |B - A| + 2|A \cap B|.$ Hence $|A \cup B| = |A| + |B| - |A \cap B|.$

Derivation of the three-set formula

Apply the two-set formula to $A$ and $B \cup C$: $|A \cup B \cup C| = |A| + |B \cup C| - |A \cap (B \cup C)|.$ Expand $|B \cup C| = |B| + |C| - |B \cap C|$ and use the distributive law $A \cap (B \cup C) = (A \cap B) \cup (A \cap C),$ so $|A \cap (B \cup C)| = |A \cap B| + |A \cap C| - |(A \cap B) \cap (A \cap C)|= |A \cap B| + |A \cap C| - |A \cap B \cap C|.$ Substitute: $|A \cup B \cup C| = |A| + |B| + |C| - |B \cap C| - |A \cap B| - |A \cap C| + |A \cap B \cap C|. \qed$

A working strategy

For a survey problem with two groups, three groups, or "at least one / exactly one":

1. Read carefully and label each given count as $|A|, |B|, |A \cap B|, |U|$ etc.
2. Draw a Venn diagram and fill in the innermost region first ($A \cap B \cap C$). Then work outwards using subtraction.
3. For "at least one of $A, B$": $|A \cup B|$. For "in $A$ only": $|A - B| = |A| - |A \cap B|$. For "neither": $|U| - |A \cup B|$.
4. For "exactly one": $|A \cup B| - |A \cap B|$ (or for three sets, sum the three single-only regions).

Worked examples

Example 1. In a class of $50$ students, $30$ play cricket, $20$ play football, and $10$ play both. How many play at least one of the two? How many play neither?

$|A \cup B| = 30 + 20 - 10 = 40$. Neither: $50 - 40 = 10$.

Example 2. Of $100$ students, $60$ like Maths, $50$ like Science, and $20$ like both. How many like Maths only? Only Science? Neither?

Maths only: $60 - 20 = 40$. Science only: $50 - 20 = 30$. At least one: $60 + 50 - 20 = 90$. Neither: $100 - 90 = 10$.

Example 3. In a survey of $200$ people, $130$ read newspaper $A$, $100$ read $B$, and $60$ read both. How many read exactly one?

Read at least one: $130 + 100 - 60 = 170$. Read both: $60$. Exactly one: $170 - 60 = 110$.

Example 4 (three sets). In a group of $400$ students, $100$ study Physics, $80$ Chemistry and $60$ Maths. $30$ study Physics and Chemistry, $25$ Chemistry and Maths, $20$ Physics and Maths, and $10$ study all three. How many study at least one subject? How many study none?

$|P \cup C \cup M| = 100 + 80 + 60 - 30 - 25 - 20 + 10 = 175$. None: $400 - 175 = 225$.

Example 5 (harder). In a survey, $70$ people speak Hindi, $50$ speak English, $40$ speak Tamil. $25$ speak Hindi and English, $20$ English and Tamil, $15$ Hindi and Tamil, and $10$ all three. How many speak (a) at least one of the three languages? (b) exactly one of the three? (c) exactly two of the three?

(a) $|H \cup E \cup T| = 70 + 50 + 40 - 25 - 20 - 15 + 10 = 110$.

(b) Region counts (using Venn diagram, working inwards–outwards):

• All three: $10$.
• $H \cap E$ only: $25 - 10 = 15$. Similarly $E \cap T$ only: $20 - 10 = 10$. $H \cap T$ only: $15 - 10 = 5$.
• Hindi only: $70 - 15 - 5 - 10 = 40$. English only: $50 - 15 - 10 - 10 = 15$. Tamil only: $40 - 10 - 5 - 10 = 15$.

Exactly one: $40 + 15 + 15 = 70$.

(c) Exactly two: $15 + 10 + 5 = 30$.

Sanity check: $70 + 30 + 10 = 110$, matching part (a).

Try it yourself

1. In a class of $60$, $35$ play hockey, $25$ play badminton, $15$ play both. How many play at least one? Neither?
2. Out of $1000$ voters, $700$ favour candidate $A$, $500$ favour $B$, $300$ favour both. How many favour neither?
3. In a group of $80$, $50$ like tea, $40$ like coffee, $20$ like both. Find the number who like only tea, only coffee, neither.
4. $|A| = 30$, $|B| = 25$, $|A \cup B| = 45$. Find $|A \cap B|$.
5. $|U| = 100$, $|A| = 50$, $|B| = 30$, $|A \cap B| = 10$. Find $|A' \cap B'|$.
6. In a group of $700$, $200$ like apples, $150$ like bananas, $100$ like cherries. $40$ like apples and bananas, $30$ bananas and cherries, $25$ apples and cherries, $10$ all three. Find at least one and exactly one.
7. In a survey of $60$ students, $30$ play cricket, $25$ tennis, $20$ both. How many play exactly one game?
8. Of $400$ households, $250$ subscribe to newspaper $X$, $180$ to $Y$, $80$ to both. Find: (a) at least one; (b) exactly one; (c) neither.
9. In a class, every student studies at least one of Maths, Physics or Chemistry. $25$ study Maths, $24$ Physics, $19$ Chemistry, $10$ Maths and Physics, $7$ Physics and Chemistry, $5$ Maths and Chemistry, $3$ all three. How many students are in the class?
10. Of $200$ students, $80$ play football, $60$ cricket, $50$ basketball. $30$ play football and cricket, $20$ cricket and basketball, $25$ football and basketball, $15$ all three. How many play none?
11. Show using inclusion–exclusion: if $|A| = 40$, $|B| = 35$, $|A \cap B| = 10$, then $|A \triangle B| = 55$.
12. In a class of $30$, $20$ passed Maths, $15$ passed Science, and $5$ failed both. How many passed both?

Pitfalls / Tricks

• Always check that $|A \cap B| \le \min(|A|, |B|)$ and $|A \cup B| \le |A| + |B|$. If your computed values violate either, you have an arithmetic error or a misread of the problem.
• Fill the Venn diagram from the innermost region outwards. Putting numbers in the wrong region is the single biggest source of mistakes.
• "At least one" $\ne$ "exactly one". Read the question precisely.
• Insight. Inclusion–exclusion is the same formula in counting and probability , replacing $|\cdot|$ by $P(\cdot)$ gives the addition rule for probability. We will meet it again in Chapter 14.