4.3. Find the number of non-empty subsets of the set $\{1,2, \ldots, 20\}$ in which the number of elements does not exceed the smallest of them.
4.3. Answer: $17711=F_{21}$ (21st Fibonacci number).

It is clear that the number of subsets of the required type, containing exactly $m$ elements, is $C_{21-m}^{m}$. Then the desired number is $-S_{21}$, where $S_{n}=\sum_{k+m=n} C_{k}^{m}$. Note that from the equality $C_{k}^{m}=C_{k-1}^{m}+C_{k-1}^{m-1}$ it follows that $S_{n}=S_{n-1}+S_{n-2}$. Since $S_{0}=S_{1}=1$, then $S_{n}$ is the $n$-th Fibonacci number.