$31 \cdot 33$ The number of ordered pairs of positive integers $(m, n)$ that are solutions to the equation $\frac{4}{m}+\frac{2}{n}=1$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(E) more than 4 .
(44th American High School Mathematics Examination, 1993)
[Solution] Since $m$ and $n$ are both positive integers, from the problem statement we know that $n>2, m>4$.
Since $\frac{4}{m}+\frac{2}{n}=1$ is equivalent to $(m-4)(n-2)=8$, we only need to find all ways to express 8 as the product of positive integers.
That is, $1 \cdot 8, 2 \cdot 4, 4 \cdot 2, 8 \cdot 1$ correspond to four sets of solutions
$$
(m, n)=(5,10),(6,6),(8,4),(12,3) \text {. }
$$

Therefore, the answer is $(D)$.