Polya's error problem

This is a problem I found in the preface of Dueling Idiots book. The problem is if Person A can detect "a" errors and Person B can detect "b" errors independent of other person and in common they detect "c" errors. What are the number of errors they missed.

Let there are "n" errors in total. $P(A) = \frac{a}{n}, P(B)=\frac{b}{n}$ then $P(A \cap B)= P(A) \times P(B)=\frac{a}{n}\times \frac{b}{n}= \frac{ab}{n^2}$. Since they have "c" errors in common which means $P(A \cap B)= \frac{c}{n}$. From these two equations we get $latex \frac{ab}{n^2}=\frac{c}{n} \Rightarrow n=\frac{ab}{c}$. Thus number of errors that are not found by both are $n -(a+b-c) = \frac{ab}{c}-(a+b-c)=\frac{ab-(ac+bc-c^2)}{c}=\frac{ab-ac-bc+c^2}{c}=\frac{a(b-c)-c(b-c)}{c}=\frac{(a-c)(b-c)}{c}$