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Exercise 6.2 · Q11

Q.If each user on a computer system has a password which is eight characters long where each character is an upper case letter or a digit. Each password must contain at least one digit. How many password are possible.

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Total 8-character passwords from 36 symbols (26 letters + 10 digits) minus those using letters only (no digit) gives 368−268=2,612,282,842,88036^8-26^8=2{,}612{,}282{,}842{,}880 valid passwords.

[!FORMULA] By the multiplication principle, the number of 8-character strings from an alphabet of size nn is n8n^{8}. Using the complement rule, (passwords with at least one digit) == (all possible passwords) −- (passwords with no digit at all, i.e. all-letter passwords).

  1. Each character is an upper-case letter (26 choices) or a digit (10 choices), so each of the 8 positions has 26+10=3626+10=36 possible characters.

  2. Total passwords (no restriction): 36836^8 (each of the 8 positions filled independently from 36 symbols).

  3. Compute 36836^8: 362=129636^2=1296, 364=12962=1,679,61636^4=1296^2=1{,}679{,}616, 368=1,679,6162=2,821,109,907,45636^8=1{,}679{,}616^2=2{,}821{,}109{,}907{,}456.

  4. Passwords with NO digit (all 8 characters are upper-case letters, violating "at least one digit"): 26826^8 (each position chosen from 26 letters).

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