Example 3-20 $n=48$, find the number of positive integers less than 48 and coprime to 48.
Due to $\sqrt{48} \leqslant 7$, any composite number less than 48 must have $2, 3,$ or $5$ as a factor.
Let $A_{1}$: the subset of numbers less than 48 that have 2 as a factor
$A_{2}$: the subset of numbers less than 48 that have 3 as a factor
$A_{3}$: the subset of numbers less than 48 that have 5 as a factor
$$
\begin{array}{l}
\phi(n)=\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3}\right| \\
=N-\left[\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|\right] \\
+\left[\left|A_{1} \cap A_{2}\right|+\left|A_{1} \cap A_{3}\right|+\left|A_{2} \cap A_{3}\right|\right]+\left|A_{1} \cap A_{2} \cap A_{3}\right| \\
\quad\left|A_{1}\right|=\left\lfloor\frac{48}{2}\right\rfloor=24, \quad\left|A_{2}\right|=\left\lfloor\frac{48}{3}\right\rfloor=16, \\
\quad\left|A_{3}\right|=\left\lfloor\frac{48}{5}\right\rfloor=9, \quad\left|A_{1} \cap A_{2}\right|=\left\lfloor\frac{48}{6}\right\rfloor=8, \\
\quad\left|A_{1} \cap A_{3}\right|=\left\lfloor\frac{48}{10}\right\rfloor=4, \quad\left|A_{2} \cap A_{3}\right|=3, \\
\quad\left|A_{1} \cap A_{2} \cap A_{3}\right|=\left\lfloor\frac{48}{30}\right\rfloor=1
\end{array}
$$

Therefore,
$$
\begin{aligned}
\phi(n) & =48-(24+16+9)+(8+4+3)-1 \\
& =48-49+15-1=13
\end{aligned}
$$

Thus, there are 13 numbers that satisfy the condition: $5,7,11,13,17,19,23,29,31,37,41,43,47$.