Let $X$ be a set and let $\mathcal{P}=\{P_1,P_2,\dots,P_m\}$ be a family of properties. Then for each subset $S\subseteq [m]\text{,}$ let $N(S)$ denote the number of elements of $X$ which satisfy property $P_i$ for all $i\in S\text{.}$ Note that if $S=\emptyset\text{,}$ then $N(S)=|X|\text{,}$ as every element of $X$ satisfies every property in $S$ (which contains no actual properties).
Returning for a moment to Example 7.1 with $P_1$ being “is a computer science major” and $P_2$ being “is male,” we note that $N(\{1\})=47\text{,}$ since there are $47$ computer science majors in the class. Also, $N(\{2\})=51$ since $51$ of the students are male. Finally, $N(\{1,2\})=45$ since there are $45$ male computer science majors in the class.
In the examples of the previous section, we subtracted off $N(S)$ for the sets $S$ of size $1$ and then added back $N(S)$ for the set of properties of size $2\text{,}$ since we'd subtracted the number of things with both properties (male computer science majors or solutions with both $x_3>7$ and $x_4'>8$) twice. Symbolically, we determined that the number of objects satisfying none of the properties was \begin{equation*} N(\emptyset) - N(\{1\}) - N(\{2\}) + N(\{1,2\}). \end{equation*}
Suppose that we had three properties $P_1,P_2\text{,}$ and $P_3\text{.}$ How would we count the number of objects satisfying none of the properties? As before, we start by subtracting for each of $P_1\text{,}$ $P_2\text{,}$ and $P_3\text{.}$ Now we have removed the objects satisfying both $P_1$ and $P_2$ twice, so we must add back $N(\{1,2\})\text{.}$ similarly, we must do this for the objects satisfying both $P_2$ and $P_3$ and both $P_1$ and $P_3\text{.}$ Now let's think about the objects satisfying all three properties. They're counted in $N(\emptyset)\text{,}$ eliminated three times by the $N(\{i\})$ terms, and added back three times by the $N(\{i,j\})$ terms. Thus, they're still being counted! Thus, we must yet subtract $N(\{1,2,3\})$ to get the desired number: \begin{equation*} N(\emptyset) - N(\{1\}) - N(\{2\}) - N(\{3\}) + N(\{1,2\}) + N(\{2,3\}) + N(\{1,3\}) - N(\{1,2,3\}). \end{equation*} We can generalize this as the following theorem: