Where is the logical flaw in my combinations answer? I encountered the following problem in DeGroot and Schervish's Probability and Statistics:

Selecting Baked Goods. You go to a bakery to select some baked goods for a dinner party. You need to choose a total of $12$ items. The baker has seven different types of items from which to choose, with lots of each type available. How many different boxfuls of $12$ items are possible for you to choose? Here we wlil not distinguish the same collection of $12$ items arranged in different orders in the box. This is an example of unordered sampling with replacement because we can (indeed we must) choose the same type of item more than once, but we are not distinguishing the same items in different orders. There are $\binom{7+12-1}{12} = 18{,}564$ different boxfuls.

My impression would be that for each of the $12$ spots, any one of the types of items could be chosen, i.e. $7$ to the $12$th power. Then, we would just divide out the different orderings of the same group, leading to the answer $7^{12} \,/\,12!$.
What's wrong with my intuition?
 A: Note that $7^{12}$ cannot possibly be a multiple of $12!$: $7^{12}$ is odd, and $12!$ is even. Thus, $\frac{7^{12}}{12!}$ clearly cannot be right. The problem is that the amount of overcounting in $7^{12}$ depends on the specific quantities of different types that you’ve chosen (and is never as large as $12!$).
Let’s look at a similar problem with smaller numbers: three types are available, $A,B$, and $C$, and you want to choose $5$ items. In the $3^5$ calculation the combination of one $A$ and four $B$s is counted $5$ times: $ABBBB,BABBB,BBABB,BBBAB$, and $BBBBA$. The combination of one $A$, two $B$s, and two $C$s, on the other hand, is counted $30$ times:
$$\begin{align*}
&ABBCC,ABCBC,ABCCB,ACBBC,ACBCB,ACCBB\\
&BABCC,BACBC,BACCB,CABBC,CABCB,CACBB\\
&BBACC,BCABC,BCACB,CBABC,CBACB,CCABB\\
&BBCAC,BCBAC,BCCAB,CBBAC,CBCAB,CCBAB\\
&BBCCA,BCBCA,BCCBA,CBBCA,CBCBA,CCBBA\\
\end{align*}$$
Thus, no there is no single number by which you can divide to get rid of the overcounting.
A: This is a "stars and bars" problem
Here is a representation of one of the choices you could have made.
⋆⋆|⋆||⋆⋆⋆|⋆⋆⋆⋆||⋆⋆
We are allocating 12 stars representing baked goods across 7 bins.  We only need 6 bars to define the 7 bins.
Your total number of possible orders is the same question of the number of ways to arrange 6 stars and 12 bars.${18\choose6}={18\choose12}$
So what do you have?
$7^{12}$ would we the first pastry selected is one of 7 varieties, and the 2nd pastry selected is on of 7...and so that would be the number of selections that could be made if order matters.
over 12!.  And then we scramble the order.
This is too much scrambling.
How to explain it....
Lets order the scenario above.
12|3||456|789A||BC
If order doesn't matter,  Moving items between bins would be a different a scenario we need to account for.  Swapping items inside a bin is not.
Good luck. 
A: The problem is that not all permutations of an ordered selection will be distinct. Suppose the selection has 12 bagels, one in each slot. Obviously any permutation of this selection leaves it exactly the same – 12 bagels. If I divided by $12!$ I would be treating all these permutations that yield the same result as different which I do not want. Therefore I have to use multiset/stars-and-bars arguments for this problem.
Then again, there are only 7 types of bread, so we must have at least two of some type. There will always be permutations that preserve the order of bread types, and these must be counted only once per distinct order.
A: Let's just say that each of the 7 types of items we can give a letter: $ABCDEFG$
When you create an order, it could be something like this:
$AABBCCDDEEFF$
That's $12$ items, but clearly there are not $12!$ ways to arrange these items. If I switched the order of the first two $A$'s, you couldn't tell. There are way fewer ways to order these items, and it actually depends on how many of each item you have in your order.
On top of that, a quick look tells you $\frac{7^{12}}{12!}$ isn't even an integer, which should indicate you probably made a mistake somewhere. ($12!$ has factors that are not $7$)
I think the biggest problem people have when they start learning combinatorics is that they forget that at the end of the day, you're trying to count stuff. Make sure that what you're counting lines up with how you're counting. Draw out the cases and make sure you didn't leave out cases or count extraneously. The whole point is to take shortcuts, but you wouldn't leap into a pool without first checking if the water was safe.
A: The problem is with the different orderings of the same group, e.g. if you take 12 times a donout, there is only one way to do so, not 12! ways to do so.
Instead you could try to count them like that: Set the 12 things in row ingoring what exactly they are: $$\circ \circ \dots \circ$$ Now to interpret them as different stuff, you add seperators, meaning everything before the first seperator is a brezel, everything between the first and second seperator is a donout and so on e.g. $$\circ \circ | | \circ \circ \circ| \circ \circ | \circ \circ | \circ \circ \circ|$$ means 2 bretzels, no donout, 3 of the third kind, 2 of fourth and fifth kind, 3 of sixth kind and none of seventh kind. Now count on how many ways you can set those seperators. Hope that helps.
