Suppose that a password for a computer system must have at least 6, but no more than 9 characters, where each character in the password is a lowercase English letter, or an uppercase English letter, or a digit, or one of the five special characters *, <, >, !, and #. (a) How many different passwords are available for this computer system? (b) How many of these passwords contain at least one occurrence of at least one of the five special characters? (c) Using your answer to part (b), determine how long it takes a hacker to try every possible password, assuming that it takes one microsecond for a hacker to check each possible password.

On each place of the password we can place 26(lowercase letters) + 26(uppercase letters) + 10(digits) + 5(special characters) = 67 characters totally

(a) For 6-digit password: $67^6$

For 7-digit password: $67^7$

For 8-digit password: $67^8$

For 9-digit password: $67^9$

Totally $67^6+67^7+67^8+67^9 \approx 2.762*10^{16}$ different passwords available

(b) We can find this amount as total amount of passwords minus anount of passwords with no special characters. A password has no special characters means on each place we can put 67 - 5 = 62 characters.

For 6-digit password: $67^6-62^6$

For 6-digit password: $67^7-62^7$

For 6-digit password: $67^8-62^8$

For 6-digit password: $67^9-62^9$

Totally $67^6+67^7+67^8+67^9-62^6-62^7-62^8-62^9 \approx 2.762*10^{16}-1.376*10^{16}=1.368*10^{16}$ different passwords with at least one occurance of at least one special characters available

(c) It will take $1.368*10^{16}$ microseconds, or ${\frac {1.368*10^{16}} {1000*60*60*24*365}}=433790$ years (433490 years if take leap years into account). All the calculations are, of course, has some inaccuracy.