4. The number of ways to choose 5 different and non-adjacent numbers from the set $\{1,2, \cdots, 25\}$ is $\qquad$ kinds.
4.20349 .

Let $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ be a subset of 5 non-adjacent numbers from $\{1,2, \cdots, 25\}$, and
$a_{i+1}-a_{i} \geqslant 2(i=1,2,3,4)$.
Let $a_{i+1}^{\prime}-a_{i}^{\prime}=a_{i+1}-a_{i}-1$, i.e.,
$\left(a_{i+1}-a_{i+1}^{\prime}\right)-\left(a_{i}-a_{i}^{\prime}\right)=1$.
Take $a_{1}^{\prime}=a_{1}$, then $a_{i}^{\prime}=a_{i}-i+1$.
Clearly, $a_{i+1}^{\prime}-a_{i}^{\prime}=a_{i+1}-a_{i}-1 \geqslant 1$, and
$$
1 \leqslant a_{1}^{\prime}<a_{2}^{\prime}<a_{3}^{\prime}<a_{4}^{\prime}<a_{5}^{\prime} \leqslant 21 \text {. }
$$

Thus, the solution is $\mathrm{C}_{21}^{5}=20349$.