find number of unordered subsets of multiset From a multiset $S = \{a, a, a, a, b, b, b, c, c\}$, how many ways are there to pick an unordered group of $3$ objects? (Objects of the same letter are identical.)
For example, in a set $\{m, n, n\}$, there are $2$ ways to pick a group of $2$ objects: $\{m, n\}$ and $\{n, n\}$.
I tried stars and bars, which didn't work the way I did it. There are varying possibilities of the amount of repetitions in the selected multiset, so the only way I can think of would be casework (which seems tedious). Is there a "smart way" to solve this problem that doesn't involve bashing?
 A: Lets make use of generating fucntion.
Generating function for $a's = 1 +x+ x^2 +x^3 +x^4$
Generating function for $b's = 1 +x+ x^2 +x^3$
Generating function for $c's = 1 +x+ x^2 $
Now , we should find the coeffiecient of $[x^3]$ in the expansion of them such that $(1 +x+ x^2 +x^3 +x^4)(1 +x+ x^2 +x^3)(1 +x+ x^2)$ , https://www.wolframalpha.com/input/?i=expanded+form+of+%281+%2Bx%2B+x%5E2+%2Bx%5E3+%2Bx%5E4%29%281+%2Bx%2B+x%5E2+%2Bx%5E3%29%281+%2Bx%2B+x%5E2%29
Answer is $9$
A: Consider the combination of $3$ stars and $2$ bars.
Suppose the number of stars to the left of the first bar give the number of solution for $a$, the number of stars to the left of the second bar give the number of solution for $b$, the number of stars to the right of the second bar give the number of solution for $c$.
(For example the following arrangement  **||* $~~$ gives the solution $a=2,b=0,c=1$. )
So there is ${5\choose 2}=10$ ways for the unordered group of $3$ objects if there is no restrictions. If the restrictions are applied then we can't choose $\{c,c,c\}$, so there will be a total of $9$ ways.
