9.23 How many four-digit numbers, composed of the digits $0,1,2,3,4,5$, contain the digit 3 (the digits in the numbers do not repeat)?
9.23 From the six digits $0,1,2,3,4,5$, one can form $A_{6}^{4}$ different four-digit combinations (including those that start with the digit 0). Among these combinations, there are $A_{5}^{3}$ combinations with zero as the first digit, so we get $A_{6}^{4}-A_{5}^{3}$ four-digit numbers composed of the given six digits. Now, from this set, we need to exclude all four-digit numbers that do not contain the digit 3, i.e., those consisting of the five digits 0,1, $2,4,5$. Reasoning as before, we conclude that there are $A_{5}^{4}-A_{4}^{3}$ four-digit numbers in which the digit 3 is absent. Therefore, the desired quantity is

$$
\begin{aligned}
& \left(A_{6}^{4}-A_{5}^{3}\right)-\left(A_{5}^{4}-A_{4}^{3}\right)= \\
& =(6 \cdot 5 \cdot 4 \cdot 3-5 \cdot 4 \cdot 3)-(5 \cdot 4 \cdot 3 \cdot 2-4 \cdot 3 \cdot 2)= \\
& =5 \cdot 60-4 \cdot 24=204
\end{aligned}
$$

Answer: 204 numbers.