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

**General Formula (Principle of Inclusion-Exclusion):** $$\left|\bigcup_{i=1}^{n} A_i\ ight| = \sum_{i} |A_i| - \sum_{i<j} |A_i \cap A_j| + \sum_{i<j<k} |A_i \cap A_j \cap A_k| - \cdots + (-1)^{n+1} |A_1 \cap \cdots \cap A_n|$$ **Example: Numbers 1-100 divisible by 2, 3, or 5.** Let $A_2, A_3, A_5$ be numbers divisible by 2, 3, 5 respectively. \begin{align*} |A_2| &= 50, \quad |A_3| = 33, \quad |A_5| = 20 \\ |A_2 \cap A_3| &= \lfloor 100/6 \ floor = 16 \\ |A_2 \cap A_5| &= \lfloor 100/10 \ floor = 10 \\ |A_3 \cap A_5| &= \lfloor 100/15 \ floor = 6 \\ |A_2 \cap A_3 \cap A_5| &= \lfloor 100/30 \ floor = 3 \end{align*} $$|A_2 \cup A_3 \cup A_5| = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74$$ **Elements in exactly one set:** Use the formula $|\text{exactly one}| = |A| + |B| + |C| - 2(|A \cap B| + |A \cap C| + |B \cap C|) + 3|A \cap B \cap C|$. For our example: $50 + 33 + 20 - 2(16 + 10 + 6) + 3(3) = 103 - 64 + 9 = 48$ numbers divisible by exactly one of 2, 3, or 5.