# How many combinations of passwords can be created according to these conditions

Suppose that a password must have at least $8$, but not more than $10$ characters, where each character in the password is either one of the $26$ lower case English letters, or one of the $10$ digits, or one of the $6$ special characters $∗, \gt, \lt, !, +, =$.

$(i)$ How many different passwords are there?

$(ii)$ How many of these passwords contain at least one occurrence of at least one of the special characters?

For part $(i)$, I think it's a total of $42$ characters therefore $42\choose10$ (no. of $10$ character passwords) minus $42\choose8$ (no. of $8$ character passwords) . For part $(ii)$, I've no idea.

I'd appreciate a bit of guidance.

• Apologies, I shall rephrase Oct 16, 2015 at 18:02
• Must the passwords have all distinct characters? Oct 16, 2015 at 18:07
• I'm not too sure, I don't think so, the question isn't clear about this Oct 16, 2015 at 18:16

i) $42^8+42^9+42^{10}$
ii)$[42^8+42^9+42^{10}]-[36^8+36^9+36^{10}]$ (The total number of possible password minus those that contains none of the special characters)
• let A be the set of all the password, let B be the set of all the password that contains at least one special character. Then $B=A-B^c$ Oct 16, 2015 at 18:10