Calculate password combination with password policy

Question

Assume user must choose 8 character password using:

lowercase letters a-z (size=26)
uppercase letters A-Z (size=26)
digits 0-9 (size=10)
special characters/symbols (size=33)

policy 1) user can choose password freely

my answer: $95^8$ combinations, since 26+26+10+33 = 95.

policy 2) The password must at least have one digit or at least one special character.

my answer: number of passwords with at least one digit = $10^8$

number of passwords with at least one special character = $33^8$

number of passwords with at least one digit or at least one special character equals to $10^8$ + $33^8$

are my calculations correct or I am making some stupid mistake?

@miracle173 - of course, I expressed my thoughts wrong on 'paper', corrected. — Leonardo, Nov 16 '16 at 14:31
Could we reach the same answer by "constructing" a valid password for policy 2? That is, picking any "digit or special" (33+10), then placing it in a position in the password (8 possibilities), then picking all the other characters ($95^7$ possibilities)? — Riccardo Orlando, Nov 16 '16 at 14:42
Does this answer your question? How many ways can you create a password of 10 characters long that has at least one lowercase letter (a-z) and at least one number ($0-9$)? — Parcly Taxel, Feb 26 '23 at 08:31

score 2 · Accepted Answer · answered Nov 16 '16 at 14:35

For your second part, number of valid passwords will be the total number possible minus the invalid ones.

Total number = $(26+26+10+33)^8 = 95^8 $

Invalid ones are those which contain neither a digit nor special character. Thus available characters are only upper and lower case alphabets. Invalid = $(26+26)^8 = 52^8 $

Thus acceptable passwords are: $95^8 - 52^8$

score 1 · Answer 2 · answered Nov 16 '16 at 14:35

1

$10^8$ is the number of passwords composed of only digits, not the number with at least one digit. The number with at least one digit is $95^8-85^8$ as we subtract all those composed of non-digits. Can you extend this?

answered Nov 16 '16 at 14:35

Ross Millikan

374,822

yep, understood – Leonardo Nov 16 '16 at 14:40

score 0 · Answer 3 · answered Nov 16 '16 at 14:46

You've made several mistakes:

Number of passwords with at least one digit is $10^8$.

No, it is $(26+26+10+33)^8-(26+26+33)^8$, or simply $95^8-85^8$.

Number of passwords with at least one special character is $33^8$.

No, it is $(26+26+10+33)^8-(26+26+10)^8$, or simply $95^8-62^8$.

Number of passwords with at least one digit or one special character is $10^8+33^8$.

No, it is $(26+26+10+33)^8-(26+26)^8$, or simply $95^8-52^8$.

Calculate password combination with password policy

3 Answers3