Question

A bank pin is a string of four digits, each digit 0-9. How many choices are there for a pin if the last digit must be odd and all the digits must be different from each other?.

Answer 1

Explanation:

Since the last digit of the pin must be odd, there are 5 choices for the last digit (1, 3, 5, 7, or 9).

For the first digit, there are 9 choices (any digit except 0 or the digit chosen for the last digit).

For the second digit, there are 8 choices (any digit except 0, the digit chosen for the last digit, or the digit chosen for the first digit).

For the third digit, there are 7 choices (any digit except 0, the digit chosen for the last digit, the digit chosen for the first digit, or the digit chosen for the second digit).

Therefore, the total number of choices for the pin is:

5 × 9 × 8 × 7 = 2520

So there are 2520 possible bank pins that meet the specified criteria.