In how many different ways can one choose 5 numbers from $\{1,2, \ldots, 18\}$, such that no two of these numbers are consecutive?
Consider the following application:

$$
\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right) \longmapsto\left(a_{1}, a_{2}+1, a_{3}+2, a_{4}+3, a_{5}+4\right)
$$

which associates with a quintuple of distinct numbers from $\{1,2, \ldots, 14\}$, ordered in ascending order, a quintuple of distinct numbers from $\{1,2, \ldots, 18\}$, also ordered in ascending order, such that no two of these numbers are consecutive. This application is clearly a bijection. The number sought is therefore equal to the number of 5-element subsets of $\{1,2, \ldots, 14\}$, the solution is $\binom{14}{5}=2002$.