$32 \cdot 40$ A fraction with a numerator of 1 and a denominator that is a natural number greater than 1 is called a unit fraction. If $\frac{1}{8}$ is expressed as the sum of two different unit fractions, then the number of all possible representation groups is
(A) 1 group.
(B) 2 groups.
(C) 3 groups.
(D) 4 groups.
(Anhui Province Anji and other counties and cities Junior High School Mathematics Competition, 1990)
[Solution 1] Clearly, if $\frac{1}{8}=\frac{1}{x}+\frac{1}{y}$, then $x>8, y>8$.
Let $x=8+m, y=8+n, \quad m \neq n, \quad m, n \in \mathbb{N}$.
That is, $\frac{1}{8}=\frac{1}{8+m}+\frac{1}{8+n}$, or $\quad \frac{1}{8}=\frac{16+m+n}{(8+m)(8+n)}$,
Simplifying the above equation gives $mn=64=1 \cdot 64=2 \cdot 32=4 \cdot 16$, that is,
$$
\begin{aligned}
\frac{1}{8} & =\frac{1}{8+1}+\frac{1}{8+64}=\frac{1}{9}+\frac{1}{72}, \\
\frac{1}{8} & =\frac{1}{8+2}+\frac{1}{8+32}=\frac{1}{10}+\frac{1}{40}. \\
\text { Also } \quad & =\frac{1}{8+4}+\frac{1}{8+16}=\frac{1}{12}+\frac{1}{24}.
\end{aligned}
$$

There are three different representations in total.
Therefore, the answer is $(C)$.
[Solution 2] Let $\frac{1}{8}=\frac{1}{x}+\frac{1}{y}$, and $x \neq y$, where $x, y \in \mathbb{N}$, and $x>1$, $y>1$.
Without loss of generality, assume $\frac{1}{x}>\frac{1}{y}$, and one of $\frac{1}{x}, \frac{1}{y}$ must be greater than $\frac{1}{16}$, then we have
$$
816 .
$$

The above equation can be transformed into $x=\frac{8 y}{y-8}$, substituting $y=9,10,11, \cdots, 15$ in sequence, we get three sets of integer solutions
$$
(x, y)=(72,9) \text { or }(40,10) \text { or }(24,12) \text {. }
$$

Therefore, the answer is $(C)$.