<s>[INST] How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$? How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$?
\begin{align*}
 A_1 & = \left\lfloor{\frac{200}{3}} \right\rfloor = 66 && \text{(divisible by $3$)}\\
 A_2 & = \left\lfloor{\frac{200}{5}} \right\rfloor = 40 && \text{(divisible by $5$)}\\
 A_3 & = \left\lfloor{\frac{200}{2}} \right\rfloor = 100 && \text{(odd)}\\
| A_1 \cap A_2 | & = \left\lfloor{\frac{200}{3 \cdot 5}}\right\rfloor = 13\\
| A_1 \cap A_3 | & = \left\lfloor{\frac{200}{3 \cdot 2}}\right\rfloor = 33\\
| A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2}} \right\rfloor= 20\\
| A_1 \cap A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2 \cdot 3}}\right\rfloor = 6
\end{align*}
Therefore, by the principle exclusion inclusion theorem
 $= 66 + 40 + 100- (13 + 33 + 20) + 6 = 146$
Is this logically right?
 [/INST] In this simple case you don't really need the Inclusion/Exclusion principle $-$ you can do it 'by hand'.
First, count the even numbers between $1$ and $30$ that are divisible by $3$ or $5$: these are $6,10,12,18,20,24,$ and $30$. So there are $7$ of them. (Or you could use Inclusion/Exclusion on this part only: $\frac{30}{6}+\frac{30}{10}-\frac{30}{30}=7$.)
Next, note that this pattern repeats exactly every $30$ numbers; so between $1$ and $210$, there are $49$ of them. Subtract $204$ and $210$ to get $47$ between $1$ and $200$.
Now add the odd numbers to get $147$.
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