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Password string consisting of one or more of the 26 characters A..Z and can be of any length from $1$ to $8$ characters. How may password available to use this scenario?

  • As my calculation, it should be $2^{26}*8!$, is it correct?
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  • for one 1 char passwords, one can only put $26$ different characters
  • for one 2 char passwords, one can only put $26^2$ different characters
  • for one 3 char passwords, one can only put $26^3$ different characters
  • ...
  • for one 8 char passwords, one can only put $26^8$ different characters

In short, think each box can contain how many letters from the given alphabet and multiply the values in the boxes. Finally, sum them.

$$\sum_{i=1}^8 26^i = \frac{26^9-1}{26-1}-1$$

The name of the formula is Sum of Consecutive powers of a number;

$$\sum_{i=0}^n a^i = \dfrac{a^{n+1} -1}{a-1}$$

and $-1$ since $i=0$ is not a case here.


Update : a more general case;

Let assume that you want to build a web application and want to determine the security level of the passwords. Here a calculator;

  • Assume that the alphabet has $l$ letters,
  • $m$ numerals, and
  • $p$ non-alphanumeric.

If we set the passwords to require at least;

  • $n_m$ numerals
  • $n_p$ non-alphanumeric,with
  • passords's length to $$min \geq n_k+n_a \geq 6,\text{and}$$
  • $max > min$, then

What is the password space, $\mathcal{S}$?

$$ \mathcal{S} = m^{n_m} + p^{n_p} + \sum_{i=1}^{max - ({n_m} + {n_p})} (l+m+p)^i$$

Under these assumptions, let

  • $l = 26$
  • $m = 10$
  • $p = 20$
  • $min = 6$
  • $max=20$
  • $n_m=2$
  • $n_p =1$, then we have;

    $$ \mathcal{S}_{6:20} = 10^2 + 20 + \sum_{i=1}^{17} (56)^i = 533361663473057950558105648760 $$

$ \mathcal{S}_{6:20}$ has 99 digits in binary form.

if max is;

  • $6 \text{ then } S = 178928$
  • $7 \text{ then } S = 10013424$
  • $8 \text{ then } S = 10013424$
  • $9 \text{ then } S = 560745200$
  • $10\text{ then } S = 31401724656$ is 35-bit.

WolframAlpha calculator;

m^{n_m} + p^{n_p} + sum (l+m+p)^i, i=1 to (max - (n_m + n_p))
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    $\begingroup$ can you please , add some explaination ? $\endgroup$ – uma Oct 6 '18 at 16:36
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No, it is incorrect. As I assume this is homework. I won't supply a full answer.

$2^{26}$ suggests 26 binary choices. But that is not the case.
$8!$ would suggest an arbitrary ordering of 8 characters.

What you have is the 8 different lengths, you can sum the number of combinations for each length.

Note letters can repeat themselves; you can select letters independently.

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  • $\begingroup$ this is not a homework . but it was question , i had read in past paper. i have to verify my answer is correct or wrong and build some discussion. this is more related to math,that is why , i expect some help . :) $\endgroup$ – uma Oct 6 '18 at 16:46
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    $\begingroup$ math.stackexchange.com/questions/739874/… $\endgroup$ – kelalaka Oct 6 '18 at 16:47
  • $\begingroup$ @kelalaka thank you.i think that is better place to this question. $\endgroup$ – uma Oct 6 '18 at 16:49

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