Venn diagrams
Draw the Venn diagram of $\sim(A \cap B)$.
Solution.
Draw the Venn diagram of $A\cup (\sim B)$.
Solution.
Draw the Venn diagram of $A\cap (B\cup C)$.
Solution.
Draw the Venn diagram of $(A-B)\cap C$.
Solution.
Draw the Venn diagram of $(B\cup C)-A$.
Solution.
Put the following Venn diagram into set notation.
Solution. $(\sim A)\cap B\cap (\sim C)$
Put the following Venn diagram into set notation.
Solution 1. $(A\cup B)-C$
Solution 2. Do disjunctive normal form. We take the union of the three shaded regions (left, middle, right), so $$ (A\;\cap \sim B\;\cap \sim C)\cup(A\cap B\;\cap \sim C)\cup (\sim A\;\cap \sim B\;\cap \sim C). $$
Counting with inclusion-exclusion and with complements
Find the cardinality of $\{1, 2, 3,\dots, 720\}-A$ where $$A=\{n\in \mathbb{Z}:n\equiv 0\;(\bmod\;8)\}.$$
Answer:Solution. To count those numbers $1,2,\dots,720$ that are not divisible by $8$, it is easier to count those numbers that are divisible by $8$.
We now want to omit elements of $A$, which have the form $n=8q$, in the range $1\le 8q\le 720$. Dividing by $8$ gives
$$ \frac18\le q\le 90 $$So we must omit the values $q=1,2,\dots,90$ which amounts to omitting $90$ elements of $A$ from the set $\{1,2,3,\dots,720\}$.
Therefore $\{1, 2, 3,\dots, 720\}-A$ has $720-90=630$ elements.
Note: A brute-force check in Python shows our answer is correct.
> # range(1,721) lists out 1, 2, 3, ..., 720
> x = [i for i in range(1, 721) if i%8 != 0]
> len(x)
How many natural numbers from 0 to 100 have no 7 in their digits?
Answer:Solution. We count by complement. Of the 101 numbers from 0 to 100, there are 10 numbers whose units place have a 7 (i.e. 07, 17, 27, ..., 97), and there are 10 numbers whose tens place have a 7 (i.e. 70, 71, ..., 79), and exactly one number with 7 in both places (77), so 10+10-1 = 19 numbers have a 7 in some of their digits. Therefore 101-19 = 82 numbers from 0 to 100 have no 7's in their digits.
Let $A=\{3n:n\in \mathbb{N}\}$ and $B=\{5n:n\in \mathbb{N}\}$. Find the cardinality of the set $(A\cap [0,100])\cup (B\cap [0,100])$
Answer:Solution. Clearly $B\cap [0,100]=\{0, 5, 10,\dots,95, 100\}$ has $21$ elements.
$A\cap [0,100]=\{0, 3, 6,\dots,96, 99\}$ has $\lfloor\frac{100}{3}\rfloor-\lceil\frac{0}{3}\rceil+1=33-0+1=34$ elements, where $\lfloor x\rfloor$ rounds down the number $x$ and $\lceil x\rceil$ rounds up the number $x$.
Finally the intersection $(A\cap [0,100])\cap (B\cap [0,100])=\{0,15,30,45,60,75,90\}$ has 7 elements.
By inclusion-exclusion, the set $(A\cap [0,100])\cup (B\cap [0,100])$ has cardinality $$ 21+34-7 = 48\text{ elements.} $$
Note: A brute-force check in Python shows our answer is correct.
> x = [i for i in range(0,101) if i%3 == 0 or i%5 == 0]
> len(x)
How many numbers in $1, 2, 3,\dots, 998, 999, 1000$ are divisible by $20$, $25$, or $30$?
Answer:Solution. Let $A$, $B$, $C$ be the sets of numbers divisible by $20$, $25$, and $110$, respectively, which are in the range $1, 2, 3,\dots, 998, 999, 1000$. Then $$ |A|=\frac{1000}{20}=50,\quad |B|=\frac{1000}{25}=40,\quad |C|=\left\lfloor\frac{1000}{30}\right\rfloor=33. $$ Next, since $20=2^2\cdot 5^1$ and $25=5^2$, any number divisible by $20$ and $25$ must also be divisible by $2^2\cdot 5^2=100$, so $A\cap B$ consists of numbers in our range divisible by $100$. Likewise $A\cap C$ consists of numbers divisible by $220$, $B\cap C$ consists of numbers divisible by $550$, and $A\cap B\cap C$ consists of numbers divisible by $1100$. Therefore: $$ |A\cap B|=\frac{1000}{100}=10,\quad |A\cap C|=\left\lfloor\frac{1000}{220}\right\rfloor=4,\quad |B\cap C|=|\{550\}|=1 $$ and $$ |A\cap B\cap C|=0 $$ By inclusion-exclusion, the answer is: $$ 50+40+33-10-4+0=109 $$