Sets, Estimation & Miscellaneous

What you need to know

• Principle of Inclusion-Exclusion (PIE), two sets. (number in $A$ or $B$) $=$ (number in $A$) $+$ (number in $B$) $-$ (number in both). Those in both were counted twice, so subtract them once. Example: of $30$ students, $18$ like math, $15$ like art, and $7$ like both. Then $18+15-7=26$ like at least one, and $4$ like neither.
• PIE, three sets. Add the three groups, subtract the three overlaps of two groups, then add back the part in all three. (That middle part was added three times and subtracted three times.) Example: from $1$ to $30$, the numbers divisible by $2$, $3$, or $5$ number $15+10+6-5-3-2+1=22$.
• "Exactly one" and "exactly two" come from the regions of the Venn diagram, not from the PIE total. Example: above, $26$ like at least one subject, but only $(18-7)+(15-7)=19$ like exactly one.
• How big can the overlap be? Say there are $N$ things in all. The number in both $A$ and $B$ is at most the smaller of the two group sizes, and at least (size of $A$) $+$ (size of $B$) $-N$. If that is negative, the smallest overlap is $0$. Example: with $30$ students, $18$ in math and $15$ in art, between $18+15-30=3$ and $15$ students are in both.
• Packing problems need two things: a lower bound from items that cannot share a container, and an actual packing that matches it. Example: boxes hold $10$ kg, and the items weigh $6,6,6,6,3,3$ kg. The total is $30$ kg, but no two $6$s fit together, so at least $4$ boxes are needed. And $4$ works: $6+3$, $6+3$, $6$, $6$.

How AMC 10 tests it

• Survey problems (positions 3–12): two or three activities that overlap, plus a group that does none; one region is asked for.
• "At least how many do both?" The tool is the bound (size of $A$) $+$ (size of $B$) $-N$.
• Packing puzzles (positions 8–15): the trap is dividing the total size by the capacity and forgetting that items can't be split.
• Estimation problems (positions 1–10): round each number to one or two significant figures.

Standard approaches

1. Draw the Venn diagram. Call the innermost region (all three) $x$, and fill in the regions from the inside out.
2. Use PIE with (number in at least one group) $=N-$ (number in none).
3. For the largest or smallest possible overlap, decide which bound the question wants.
4. For packing, list which combinations fit in one container, find the bound, then build a packing that reaches it.

Worked example

In a class of $40$ students, $24$ play soccer, $20$ basketball, and $18$ tennis. Ten play soccer and basketball, $8$ soccer and tennis, $7$ basketball and tennis, and $2$ play none. How many play exactly one sport?

(A) $13$ (B) $14$ (C) $15$ (D) $16$ (E) $17$

Find the middle. Let $x$ students play all three. Since $2$ play none, $40-2=38$ play at least one. By PIE,
$$38=24+20+18-10-8-7+x=37+x,$$
so $x=1$.

Find the "exactly two" regions. The $10$ soccer-and-basketball players include the $1$ who plays all three, so $10-1=9$ play only those two. In the same way, $8-1=7$ play only soccer and tennis, and $7-1=6$ play only basketball and tennis.

Subtract. At least two sports: $9+7+6+1=23$ students. So exactly one sport: $38-23=15$. (Check: the one-sport regions hold $7$, $4$, and $4$.) The answer is $\boxed{\textbf{(C)}\ 15}$.

Pitfalls

• Reading "$10$ play soccer and basketball" as leaving out the ones who also play tennis. Those $10$ include everyone who plays both, tennis or not.
• Forgetting the "none" group: in the example, the students in at least one sport number $38$, not $40$.
• In packing, stopping at the total divided by the capacity, rounded up. Above, $30\div10=3$, but $4$ boxes are needed.
• Rounding every number down in an estimate. Example: $3.9\times4.9\times5.9$ is about $4\times5\times6=120$ (it is $112.749$), not $3\times4\times5=60$.

Traps that recur

• Forgetting that 1 through 24 are unconstrained and answering 50 (one from each of fifty imagined pairs). (2005 AMC 10B #25)
• Assuming the answer is just the 20% net change, or letting more students switch Yes-to-No than the 30% who answer No at the end. (2010 AMC 10B #12)
• Dividing total size 16.8 MB by 1.44 to get 12 disks, ignoring that files cannot be split and that 0.8 and 0.7 cannot share a disk. (2002 AMC 10A #11)
• Answering 4 from 52/12 = 4.33 rounded down, or 12 from misreading the question as the number of months. (2011 AMC 10B #11)