Question: In a certain company, employee IDs are strings of 4 digits (that may begin with a $0$).
How many IDs begin with an even digit, end with an odd digit, and have no repeated digit?
Answer:
- First digit: $5$ choices (either $0$, $2$, $4$, $6$, or $8$)
- Last digit: $5$ choices (either $1$, $3$, $5$, $7$, or $9$)
- Second digit: $8$ choices ($5+5 - 2$)
- Third digit: $7$ choices ($5+5 - 3$)
Total number of IDs = $5 \times 5 \times 8 \times 7$
Am I right?