A factory produces 10 items. In how many ways can these items be categorized into 3 quality levels? If the items are indistinguishable will the problem follow the model of dividing $10$ objects into $3$ bins, where the solution is $n+m-1$ choose $n$ so $12$ choose $10$? Or will this be more like a password problem where the solution is $3^{10}$?
What happens if it is required that there be at least $1$ object in each category?
 A: First to make it clear if the 10 objects are undistinguishable then it's just a simple stars and bars problem, while if the 10 objects are different the total number of ways is $3^{10}$ as you mentioned.
If the 10 objects are undistinguishable, then to categorize them in 3 ways we use the stars and bars formula:
$$\binom{n+k-1}{k} = \binom{3+10-1}{10} = \binom{12}{10} = 66$$
The stars and bars problem is actually finding a number of distinct ways to make $n$-tuples, such their sum is $k$, where are number are non-negative integers.
If we want to restrict them only on positive integers then the formula is:
$$\binom{k-1}{n-1} = \binom{10-1}{3-1} =\binom{9}{2} = 36$$
So there are 36 distinct ways, such that every category has at least on item.
To find out more about the principle this formula works, look at the Wikipedia page, you have a very nice exlpanation.
A: The classic stars and bars approach here is the approach to use indeed (using approach $2$, theorem 2 in the link): 
In this case, the task is equivalent to counting the number of non-zero integer solutions to the equation $x_1 + x_2 + x_3 = 10$, where $k = 10$, $n = 3$ bins. in this case:  $$\binom{n + k - 1}{k} = \binom{10 + 3 - 1}{10} = \binom {12}{10}$$
In the second case, we can use theorem one from the same problem type of the "stars and bars" sort: the count is equivalent to the number of positive integer solutions (so no bin can be empty) to in this case $x_1 + x_2 + x_3 = 10$, where $k=10$ objects and $n = 3$ bins: $$\binom{k - 1}{n - 1}\binom{10 - 1}{3-1} = \binom{9}{2}$$
