I'm trying to figure out how many possible passwords can be created from these three questions. Repetition is allowed.

  1. $8$ characters long, at least $1$ capital, at least $1$ number
  2. $8$ characters long, at least $1$ capital
  3. $8$ characters, at least $1$ capital, $1$ lowercase, $1$ number, $1$ special character ($3$ total special characters)

My rough work - I'm unsure how to do this, the at "least portion" of the question confuses me....

  1. $nCr(52,1)^6 \cdot nCr(10,1) \cdot nCr(26,1)$

  2. $nCr(52,8)$

  3. $nCr(26,1) \cdot nCr(26,1) \cdot nCr(10,1) \cdot nCr(3,1) \cdot nCr(52,1,)^4$

$52 =$ Capitals and lowercases, $10 =$ Numbers from $0-9$, $3 =$ Special Characters

  • $\begingroup$ No. $52^6\cdot 10\cdot 26$ is the number of passwords there are where very specifically the first six characters are letters, very specifically the seventh character is a number, and very specifically the eighth character is a capital letter. Your attempt for part 2 is the number of eight character strings who have no repetition and the letters very specifically occur in alphabetical order (capital>lowercase) $\endgroup$
    – JMoravitz
    May 6 '20 at 0:32

$8$ characters long, at least $1$ capital, at least $1$ number

The complement of "at least 1" is "none".

Use the principle of inclusion and exclusion.

$$\mu(\text{Any 8 Chars})-\mu(\text{No Caps})-\mu(\text{No Digits})+\mu(\text{Neither Caps nor Digits})$$

Note that $\mu(\text{Any 8 Chars})=(26+26+10+3)^8$ when there are 26 capitals, 26 lowercase, 10 digits, and 3 special characters, and repetitions are allowed.

Solve the others similarly.

  • $\begingroup$ can you please elaborate on the 'neither caps nor digits' for some reason it didnt click to me. $\endgroup$
    – adhg
    May 6 '20 at 1:33
  • 1
    $\begingroup$ Draw a Venn Diagram for $A$, strings with Any 8 Characters, and inside that set, overlapping subsets for $C_0$, strings with No Caps, and $D_0$, strings with No Digits. The intersection of these is strings with Neither Caps nor Digits (that is: 0 caps and 0 digits). We wish to measure the (probability) size of Any string except those with 0 caps or 0 digits (ie: at least 1 each).$$\lvert A\smallsetminus(C_0\cup D_0)\rvert~{=\lvert A\rvert-\lvert A\cap C_0\rvert-\lvert A\cap D_0\rvert+\lvert A\cap C_0\cap D_0\rvert\\ =\lvert A\rvert-\lvert C_0\rvert-\lvert D_0\rvert+\lvert C_0\cap D_0\rvert}$$ $\endgroup$ May 6 '20 at 2:01
  • $\begingroup$ Got it! thank you. $\endgroup$
    – adhg
    May 6 '20 at 2:05

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