How to use the inclusion-exclusion principle for counting problems?

The inclusion-exclusion principle is a way to count a union without double-counting overlaps. For three sets, the formula is: $$|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|A \cap C|-|B \cap C|+|A \cap B \cap C|.$$ For $n$ sets, the signs alternate: $$\left|\bigcup_{i=1}^{n} A_i\right| =\sum_i |A_i| -\sum_{i<j}|A_i\cap A_j| +\sum_{i<j<k}|A_i\cap A_j\cap A_k| -\cdots.$$ Now count integers from 1 to 100 divisible by 2, 3, or 5. $$|A_2|=50,\quad |A_3|=33,\quad |A_5|=20.$$ The pairwise overlaps are multiples of the least common multiples: \[ \begin{aligned} |A_2\cap A_3|&=\left\lfloor \frac{100}{6}\right\rfloor=16,\\ |A_2\cap A_5|&=\left\lfloor \frac{100}{10}\right\rfloor=10,\\ |A_3\cap A_5|&=\left\lfloor \frac{100}{15}\right\rfloor=6. \end{aligned} \] The triple overlap is: $$|A_2\cap A_3\cap A_5|=\left\lfloor \frac{100}{30}\right\rfloor=3.$$ So the number divisible by at least one of 2, 3, or 5 is: $$50+33+20-16-10-6+3=74.$$ For "exactly one" of the three sets, subtract the pairwise overlaps twice and add the triple overlap three times: $$|A|+|B|+|C|-2(|A\cap B|+|A\cap C|+|B\cap C|)+3|A\cap B\cap C|.$$ Here that gives: $$103-2(32)+3(3)=48.$$ That last formula works because an element in exactly two sets was counted twice at first and must be removed entirely, while an element in all three sets needs its own correction.