Question

A bank PIN is a string of four digits, each digit 0-9. (a) How many choices are there for a PIN if the last digit must be odd? Make sure to explain your answer. (b) 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? Make sure to explain your answer.

Answer 1

For the first part, there are 5,000 possible PIN combinations where the last digit must be odd. For the second part, where the last digit must be odd and all digits different, there are 3,600 possible combinations.

Step-by-step explanation:

The student has asked two problems related to calculating the number of combinations for a bank PIN, subject to certain conditions. Let's address each problem one by one.

Part (a)

For the first problem, the student is asked how many choices there are for a PIN if the last digit must be odd. Since a bank PIN is a string of four digits, and each digit can range from 0-9, for the first three positions, we have 10 possibilities (0-9). However, for the last position, because it must be odd, we only have 5 possibilities (1, 3, 5, 7, or 9). This results in a total of 10 x 10 x 10 x 5 = 5,000 possible combinations.

Part (b)

For the second problem, the last digit must still be odd, but now all digits must be different. The last position has 5 choices as explained before. For the first digit, we have 10 possible choices. For the second digit, we have 9 remaining choices (excluding the first digit), and for the third digit, we have 8 remaining choices (excluding the first two digits). Therefore, the total number of possible combinations is 10 x 9 x 8 x 5 = 3,600.