For example, how many ways can you choose $12$ items from $8$ item types? One possibility is that you choose every item from just one of the types.
But what if we have the restriction that we must have at least $1$ of each type?
The way I think of this is that $8$ out of our $12$ choices are locked in, so we only have freedom in choosing the remaining $4$ items, and so the answer is $8$multichoose$4$, and the calculation is carried out as $$\frac{11!}{4!7!}= 330$$
Does this make sense? Any help is appreciated!
You have to choose 4 objects out of 8 object types. I am assuming that you have atleast 5 objects of each object type. As you have already selected 1, you have 4 objects remaining of each object type. The number of ways you can do this is basically the coefficient of $x^4$ in the series $(1+x^2+x^3+...)^{8}$ by the method of generating functions. The series can be written as $(\frac{1}{1-x})^{8} $. You can get the coefficient of $x^4$ with the help of binomial theorem as $\frac{8*9*10*11}{4!}$.