How do I calculate a password space if the password only has eight alphabetic characters (not case sensitive) I read this wiki article and saw this formula for random password's information entropy. Also the wiki article mention that the symbol count for non case sensitive is 36. Is there a way I can calculate the password space with this information.
The choice of 8 fully random characters is exactly 3 bits as $\log_2(8) = 3$. So to calculate the strength in bits you simply multiply with the size of the password: e.g. 16 * 3 = 48 bits of strength for 16 characters.

To translate that to the password space - the number of possible passwords - you simply can perform $2^{48}$ which means that there are 281,474,976,710,656 possible passwords. A little known trick to get an idea of the number of decimal digits is to multiply by 3 and then divide by 10 and round up: so $\large\lceil\normalsize{48 * 3 \over {10}}\large\rceil\normalsize = 15$, i.e. above number contains 15 digits. A more direct method is to simply perform exponentiation with the password size: $8^{16}$ will give you the same large number. However, I use the trick above to perform such calculations without calculator.

Note that protection against brute force should require key stretching by e.g. using a password hash. For such kind of passwords additional protection (such as account locking) should be implemented. The system needs to deal with low entropy passwords.



The trick above depends on a little known constant: 3.3219280948873626. This is the result of $\ln(10)\over\ln(2)$, i.e. each decimal digit takes about 3.3219280948873626 bits to be represented. It's quite close to 3.3333 wich makes calculations rather efficient. The little difference makes it so that $2^{10} = 1024$, or just over a thousand (a kilo in ancient Greek).