Question

Assume that a password must be at least 8 characters long and include at least 1 digit and at least 1 special character. If people form their password by taking an English word of exactly 6 letters (assume that this password is not case-sensitive) and then adding (in either order) a digit and a special character (e.g flames#1), how many passwords would be possible? (10 points)

Answer 1

Answer:

10,932,240

Step-by-step explanation:

According to the Naspa World list American english have 16,564 6-letters words. Now about the special characters we have the next list  !"#$%&'()*+,-./:;<=>[email protected][\]^_`{|}~ and considering the space as a special character we have a total of 33 special characters. For numbers we have a total of 10 digits.

Then to know how many possibles exists we have to find how many possibles are for the last two characters then.

$33\cdot10=330$

That is the amount os possibles if always the special character go before de number, but as the number could be before the special character we have to multiply this quantity by 2.

Then we have 16,564 words for the first 6 characters and 660 options for the last two. To know the total amount of possibilities we just need to multiply this numbers, then:

$16,564\cdot660=10,932,240$