If someone say(verbally) his/her password is 12345678. Some example possible password are

"12345678" or "Onetwothreefourfivesixseveneight" Or "1twothreefourfivesixseveneight", "1twothreefourfivesixseven8" ...

Letters can be upper and lower case both. So what is the number of total possible passwords or what can be upper bound for the number of such passwords.

Some smaller cases

If they say it's "1" : 9

Then we have One, ONe, ONE, ... 8 possibilities and 1

so there are 9 passwords.

If they say it's "12": 82

then we have 8*8 only for digits case.. =64

then 12 = 1

mixed are : one2, One2 ... 8 possibilities. + 1two, 1Two .8 possibilities = 16 should be 64+2+16 = 82

We can discount far-fetched interpretations such as "444" for "three four", since this is ungrammatical.


closed as unclear what you're asking by Leucippus, N. F. Taussig, Namaste, Shailesh, NCh Feb 10 '18 at 1:37

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    $\begingroup$ This is too vague, which possibilities do you take as 'correct' and which not? $\endgroup$ – abcdef Feb 9 '18 at 15:52
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    $\begingroup$ Seems you are trying to bruteforce into someone's account. Well I think the numbers finite, but cases are way too many for anyone to answer. $\endgroup$ – King Tut Feb 9 '18 at 15:53
  • $\begingroup$ The rules aren't specified. Could I read this as $234(678)^5$ (as one $234$ five $678's$)? The english language isn't mathematically precise. $\endgroup$ – lulu Feb 9 '18 at 15:54
  • $\begingroup$ Welcome to stackexchange. You're more likely to get answers rather than downvotes or votes to close if you edit the question to show us what you tried and where you are stuck. You should at least work out the cases "12" , "123" and "1234". $\endgroup$ – Ethan Bolker Feb 9 '18 at 15:54
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    $\begingroup$ What about "won", "to", "too", "for", "ate", and so on? $\endgroup$ – Théophile Feb 9 '18 at 16:12

For each digit, you have either the digit itself, or the corresponding word. Each letter of the word can be upper or lower case, so in total there are $1 + 2^l$ possibilities for a given digit, where $l$ is the number of letters in the word.

Thus, the number of possibilities for $123456789$ is $$(1+2^3)(1+2^3)(1+2^5)\cdots(1+2^4) = 128,711,132,649.$$

I have discounted far-fetched interpretations such as "444" for "three four", since this is ungrammatical. The one exception would be "2" for "one two", as the singular here is grammatical.

  • $\begingroup$ I am not getting same number while calculating as this formula. $\endgroup$ – Saurabh Prajapati Feb 9 '18 at 16:41
  • $\begingroup$ I was calculating for 8 only .. your calculation is correct. $\endgroup$ – Saurabh Prajapati Feb 9 '18 at 16:44
  • $\begingroup$ Oh, sorry, indeed you asked for 12345678 while this is 123456789. In any case, I hope the point is clear. $\endgroup$ – Théophile Feb 9 '18 at 19:40

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