How many fractions less than 1 exist, such that the numerator and denominator are one-digit natural numbers?
For a fraction to be less than 1, the numerator must be less than the denominator. The fractions are:

- with denominator 2: $\frac{1}{2}$
- with denominator 3: $\frac{1}{3}$ and $\frac{2}{3}$
- with denominator 4: $\frac{1}{4}, \underbrace{\frac{2}{4}}_{1 / 2}$ and $\frac{3}{4}$
- with denominator 5: $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}$
- with denominator 6: $\frac{1}{6}, \underbrace{\frac{2}{6}}_{1 / 3}, \underbrace{\frac{3}{6}}_{1 / 2}, \underbrace{\frac{4}{6}}_{2 / 3}, \frac{5}{6}$
- with denominator 7: $\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$
- with denominator 8: $\frac{1}{8}, \underbrace{\frac{2}{8}}_{1 / 4}, \frac{3}{8}, \underbrace{\frac{4}{8}}_{1 / 2}, \frac{5}{8}, \underbrace{\frac{6}{8}}_{3 / 4}, \frac{7}{8}$
- with denominator 9: $\frac{1}{9}, \frac{2}{9}, \underbrace{\frac{3}{9}}_{1 / 3}, \frac{4}{9}, \frac{5}{9}, \underbrace{\frac{6}{9}}_{2 / 3}, \frac{7}{9}, \frac{8}{9}$

We then have 27 fractions.