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0 Rebecca has a combination lock with 4 numbers.
Each dial contains the numbers 0-9 inclusive.
She is trying to remember the code.
3
She knows the first number is 3, the last number is even and that the second and
third numbers are different.
How many possible different codes are there?

Verified Answer :

GinnyAnswer

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:

\(\[ \text{Number of possible codes} = (1 \text{ choice for the first digit}) \times (10 \text{ choices for the second digit}) \times (9 \text{ choices for the third digit}) \times (5 \text{ choices for the fourth digit}) \]\)

\(\[ \text{Number of possible codes} = 1 \times 10 \times 9 \times 5 = 450 \]\)

Therefore, there are 450 possible different codes for Rebecca's combination lock.

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