Let's solve this question step-by-step:
1. First Digit:
- Rebecca remembers that the first number is 3. Since this is fixed, there is only 1 choice for the first digit.
2. Second Digit:
- The second digit can be any number from 0 to 9. Therefore, there are 10 possible choices for the second digit.
3. Third Digit:
- The third digit must be different from the second digit. Since the second digit can be any number from 0 to 9, and the third digit needs to be different from it, there are 9 possible choices for the third digit (one less than the total options for the second digit to ensure it is different).
4. Fourth Digit:
- The last digit must be an even number. The digits that are even between 0 and 9 are 0, 2, 4, 6, and 8. Therefore, there are 5 possible choices for the fourth digit.
Now, we can find the total number of possible different codes by multiplying the number of choices for each position:
[tex]\[
\text{Number of possible codes} = (1 \text{ choice for the first digit}) \times (10 \text{ choices for the second digit}) \times (9 \text{ choices for the third digit}) \times (5 \text{ choices for the fourth digit})
\][/tex]
[tex]\[
\text{Number of possible codes} = 1 \times 10 \times 9 \times 5 = 450
\][/tex]
Therefore, there are 450 possible different codes for Rebecca's combination lock.
