Total password count, given length and pool size
Possible combinations (permutations) = sequenceOptions to the power of sequenceLength

Assuming a pool of 95 sequence options


sequenceLength 0 = 0 permutations (950)
sequenceLength 1 = 95 permutations (951)
sequenceLength 2 = 9025 permutations (952)
sequenceLength 3 = 857375 permutations (953)
etc


This is not limited to passwords/characters. For example, when a password is put through a key derivation algorithm, it ends up as a string of bits for which there are two options; 0 and 1.  Hence, a 256 bit length key contains 2256 permutations. 

More simply, a 4-digit numeric pincode has 10000 (104) possible permutations.

As for the second part of your question, yes; if the sequenceLength were &lt;= 8 then the total number of permutations would be 950 + 951 + 952 + 953 + 954 + 955 + 956 + 957 + 958

Of course, the above assumes that each option in the sequence can appear more than once. If that's not the case then factorials come into play.