In how many ways can you distribute 
Of course, for each object, there are 
Now, consider the same question, except, the objects are not distinct any more. Thus all that matters is how many of them go into each bin and not which ones in particular.
The answer to this can be slightly complicated to calculate, but if we simplify things a bit and look for a loose upper bound, we can argue as follows. The problem is equivalent to assigning one number between 0 and 
I am certain that I knew both these things back in high school, but it was quite recently that I realized that the first one is exponential in
Things 
If the objects aren’t distinct then the problem is the same as finding the number of non-negative integer solutions to
. This problem can again be reduced to a choosing problem to give 
.
If you continue varying the possibilities of such problems you notice that you essentially have three parameters: distinctness of bins, distinctness of balls, restrictions on the number of balls in a bin. The first two parameters can take two values, while the last parameter can take one of three values: i) at least one ball, ii) at most one ball, iii) no restriction. Finding the solution of each combinatorial problem will lead you to the twelvefold way (in your post, you found two entries in that table).
The aim, obviously, was not to enumerate entries from that table. The aim was to point out that there is an exponential gap between two of the entries.
Oh, oops. Yeah, I never noticed the dramatic difference in the order of growth in
either.