Computing the size of an m-associative cache I am trying to answer this question:

Consider a computer with a byte-addressable memory. A 40-bit memory address is divided as follows for cache processing. First, the 8 low-order bits are chopped off to expose the cache-line number. Second, the next 17 low-order bits are inspected to get the cache-container index.  Third, the remaining 15 bits are used as the cache tag. Hint: What do the direct-mapped and set-associative placement formulas have in common?

What is the cache size in bytes?

I've figured out that the cache line size is 2^8 bytes and that there are thus 2^32 cache lines in memory.
Furthermore I know that there are 2^17 cache sets in the cache.
To compute the capacity of the cache, I am using the equation

capacity = line size x #of sets x associativity

However, even using the tag field, I can't see how I can find the associativity of the cache here.
I'm not sure if the first question is actually a question or a Hint. Anyway cache size can be calculated as follows.

Memory is byte addressable. 
Masking off 8 bits for cache line, that means, cache line in 2 ^ 8 bytes.
Using 17 bits for TAG indexing, that is 2^17 cache entries.
Therefore, you have 2^8*2^17 bytes in the cache. That is 2^25 = 2^5*2^20 = 32 MB.