How many fractions less than 1 are there, in which the numerator and the denominator are single-digit positive integers?
For a fraction to be less than 1, the numerator must be less than the denominator. Eliminating repetitions, we obtain the following list.

(a) 1 fraction with denominator 2: $\frac{1}{2}$

(b) 2 fractions with denominator $3: \frac{1}{3}$ and $\frac{2}{3}$

(c) 2 fractions with denominator 4: $\frac{1}{4}, \underbrace{\frac{2}{4}}_{1 / 2}$ and $\frac{3}{4}$

(d) 4 fractions with denominator 5: $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}$ and $\frac{4}{5}$

(e) 2 fractions with denominator 6: $\frac{1}{6}, \underbrace{\frac{2}{6}}_{1 / 3}, \underbrace{\frac{3}{6}}_{1 / 2}, \underbrace{\frac{4}{6}}_{2 / 3}$ and $\frac{5}{6}$

(f) 6 fractions with denominator $7: \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}$ and $\frac{6}{7}$
(g) 4 fractions with denominator $8: \frac{1}{8}, \underbrace{\frac{2}{8}}_{1 / 4}, \frac{3}{8}, \underbrace{\frac{4}{8}}_{1 / 2}, \frac{5}{8}, \underbrace{\frac{6}{8}}_{3 / 4}$ and $\frac{7}{8}$

(h) 6 fractions with denominator 9: $\frac{1}{9}, \frac{2}{9}, \underbrace{\frac{3}{9}}_{1 / 3}, \frac{4}{9}, \frac{5}{9}, \underbrace{\frac{6}{9}}_{2 / 3}, \frac{7}{9}$ and $\frac{8}{9}$.

Thus, we have 27 of these fractions.