# What is the number of all the possible passwords when someone say(verbally) password is 12345678. [closed]

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

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, NChFeb 10 '18 at 1:37

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• This is too vague, which possibilities do you take as 'correct' and which not? – abcdef Feb 9 '18 at 15:52
• 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. – King Tut Feb 9 '18 at 15:53
• 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. – lulu Feb 9 '18 at 15:54
• 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". – Ethan Bolker Feb 9 '18 at 15:54
• What about "won", "to", "too", "for", "ate", and so on? – 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 $$(1+2^3)(1+2^3)(1+2^5)\cdots(1+2^4) = 128,711,132,649.$$