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.

  • $\begingroup$ Apologies, I shall rephrase $\endgroup$
    – aqibjr1
    Oct 16, 2015 at 18:02
  • 1
    $\begingroup$ Must the passwords have all distinct characters? $\endgroup$ Oct 16, 2015 at 18:07
  • $\begingroup$ I'm not too sure, I don't think so, the question isn't clear about this $\endgroup$
    – aqibjr1
    Oct 16, 2015 at 18:16

1 Answer 1


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)

  • $\begingroup$ 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$ $\endgroup$ Oct 16, 2015 at 18:10
  • $\begingroup$ If you wanted to select the no. of ways you can choose 8 characters from 42, would it not be 42C8? And therefore the answer be 42C8+42C9+42C10? $\endgroup$
    – aqibjr1
    Oct 16, 2015 at 18:11
  • $\begingroup$ The order matters here, and it is with repetition. $\endgroup$ Oct 16, 2015 at 18:13
  • $\begingroup$ When you say 42C8, you are assuming that the order doesn't matter, and it is without repetition. $\endgroup$ Oct 16, 2015 at 18:14
  • $\begingroup$ Ah okay, I get why repetition would be allowed, but surely the order of the characters in the password wouldn't? $\endgroup$
    – aqibjr1
    Oct 16, 2015 at 18:15

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