Brute force reshuffle

Question

Assuming an alphanumeric password of 8 characters, the number of permutations in my understanding will be.

26 lowercase letters
26 uppercase
10 digits

So, if you made a brute force attack on this password, the average number of attempts would be (62 ^ 8) / 2

However, assuming that you knew that the password was at least 4 digits and therefore excluded any attempts, the first 4 digits will not respond to the other permutations ((62 ^ 8) - (62 ^ 4)) / 2?

Am I missing something or is this the correct answer?

Answer 1

Yes, you are missing something. No, this is not the right answer :-)

, .

, , , ( , , , ).

:

(62 ^ 1) + (62 ^ 2) + (62 ^ 3) + (62 ^ 4) +
(62 ^ 5) + (62 ^ 6) + (62 ^ 7) + (62 ^ 8)

( , ).

, , , , :

                                 (62 ^ 4) +
(62 ^ 5) + (62 ^ 6) + (62 ^ 7) + (62 ^ 8)

Answer 2

8 62^8 + 62^7 + 62^6 + 62^5 + 62^4 + 62^3 + 62^2 + 62 . , , 4 , 3 - 62^8 + 62^7 + 62^6 + 62^5 + 62^4 .

Answer 3

I assume that the correct calculation of the number of possible combinations for your password will be:

(62^4) + (62^5) + (62^6) + (62^7) + (62^8)