9. (3 points) Using the digits $0, 1, 2, 3, 4$, the number of even numbers that can be formed without repeating any digit is $\qquad$.
9. (3 points) Using the digits $0, 1, 2, 3, 4$, a total of $\qquad$ 163 different even numbers without repeated digits can be formed.

【Solution】Solution: Single-digit even numbers are: 0, 2, and 4, which are 3 in total;
Two-digit even numbers: $10, 20, 30, 40, 12, 32, 42, 14, 24, 34$, there are 10 in total;
Three-digit even numbers:
When the last digit is 0, the tens and hundreds place can be selected from 4 elements, with $A_{4}^{2}=12$ possible arrangements. When the last digit is not 0, it can only be 2 or 4, the hundreds place can be selected from 3 elements, and the tens place from 3 elements, resulting in $A_{2}^{1} A_{3}^{1} A_{3}^{1}=18$ possible arrangements.

According to the principle of classification counting, there are $12+18=30$ possible results;
Four-digit even numbers:
When the last digit is 0, there are $\mathrm{A}_{4}^{3}=24$ such four-digit numbers,

When the last digit is 2 or 4, there are $2 \times C_{4}^{1} \times \mathrm{A}_{3}^{2}=36$ such four-digit numbers,
In total, there are $24+36=60$ (numbers)
Five-digit even numbers:
When the last digit is 0, there are $A_{4}^{4}=24$ such five-digit numbers,
When the last digit is 2 or 4, there are $2 \times C_{3}^{1} \times A_{3}^{3}=36$ such five-digit numbers,
Therefore, there are $24+36=60$ five-digit even numbers without repeated digits.
In total: $3+10+30+60+60=163$ (numbers);
Answer: A total of 163 different even numbers without repeated digits can be formed.
Thus, the answer is: 163.