Question

QUESTION 2. Find the number of possible choices for a 2-digit password that is greater than 19. Then find the number of possible choices for a 4-digit Personal Identification Number (PIN) if the digits cannot be repeated.

Answer 1

QUESTION 2. Find the number of possible choices for a 2-digit password that is greater than 19. Then find the number of possible choices for a 4-digit Personal Identification Number (PIN) if the digits cannot be repeated.

Solution

1) The number of possible choices for a 2-digit password that is greater than 19:

$$(A_{10}^2 + 10) - 20 = \left(\frac{10!}{(10 - 2)!} + 10\right) - 20 = \left(\frac{10!}{8!} + 10\right) - 20 = (9 * 10 + 10) - 20 = 80.$$

$A_{10}^2$ - permutations without repetitions;

10 - number of 00, 11, 22, 33, 44, 55, 66, 77, 88, 99;

20 - number of 2-digit password that smaller than 19;

Answer: The number of possible choices for a 2-digit password that is greater than 19 are 80.

2) The number of possible choices for a 4-digit Personal Identification Number (PIN) if the digits cannot be repeated:

$$A_{10}^4 = \frac{10!}{(10 - 4)!} = \frac{10!}{6!} = 7 * 8 * 9 * 10 = 5040;$$

Answer: The number of possible choices for a 4-digit Personal Identification Number (PIN) if the digits cannot be repeated are 5040.