
Number of choices
This one is from Hogg & Tanis, 1.321.
A manufacturer makes storage files, which are made up of a top, a base, and up to six storage modules. There are four bases to choose from and two tops, and the storage modules come in five different sizes.
a) How many choices are possible for a completed storage file made up of a top, a bottom, and four storage modules? The stacking order of the modules is not important. (answer: 560)
b) How many choices are possible for a completed storage file made up from a top, a bottom, and between one and six storage modules? (answer: 3688)
For both of them the number of choices is equal to 2 * 4 * {number of different module configurations} but I am having a lot of difficulty enumerating those configurations.
In part a) I finally resorted to splitting the number of configurations into how many different sizes are possible.
If all the modules are the same size, there are 5 options; if there are either two or three different sizes respectively then there are $\displaystyle 3\cdot\binom{5}{2} = 3\cdot\binom{5}{3} = 30$ options (in either case); and if there are four different sizes then there are also five options. That makes 70 possibilities for the modules, multiplied by 8 different top/bottom combinations which gives the answer.
The thing is, that took me ages and I basically had to do it by listing rather than using combinations, which is the whole point of the question. I can't apply the above properly to part b) and I shouldn't have to do it that way in any case. Can anyone help? Thanks!

Re: Number of choices
For part a), if you want to figure out the number of combinations there are for the sizes, consider this:
You have four modules, and five sizes. Each size of module is like a box. You can throw all four into a single box, or spread them out. Let a module be an x and a box be the space between vertical lines . Then, the five "sizes" could be considered  and the four modules xxxx. Now, each possible order of xxxx tells us a configuration for the modules of various sizes. This example would be one module of the first size, two of the second size, and one of the fourth. So, how many possible orders are there? Well, you have 8 slots where you are putting 4 x's and the rest of the slots get a . So that is $\displaystyle \binom{8}{4}$ different choices. So, there are $\displaystyle 2\cdot 4 \cdot \binom{8}{4} = 560$ options for part a), just as you calculated.
Now, for part b), there are
$\displaystyle \begin{align*}2\cdot 4\cdot \left[ \binom{5}{1} + \binom{6}{2} + \binom{7}{3} + \binom{8}{4} + \binom{9}{5} + \binom{10}{6} \right] \\ = 8\cdot (5 + 15 + 35 + 70 + 126 + 210) = 3688\end{align*}$ different possible configurations.

Re: Number of choices
That makes a lot of sense. Thank you :)