AHSME 1993 Problem 19

How many ordered pairs $(m,n)$ of positive integers are solutions to $$\frac{4}{m}+\frac{2}{n}=1?$$ $\text{(A) } 1 \text{(B) } 2 \text{(C) } 3 \text{(D) } 4 \text{(E) } \text{more than }6$