The notation $\left( \! \binom{k}{n} \! \right)$ stands for the number of multisets of size $n$ on $k$ symbols. The important thing to note is that the $k$ symbols are distinguishable and the $n$ "positions" in the multiset are indistinguishable (order doesn't matter).
This suggests a bijection (one-to-one correspondence) with the number of ways to put balls (indistinguishable objects) in labeled boxes (distinguishable objects). Specifically, the $k$ symbols in the multiset correspond to the boxes, and the $n$ positions in the multiset correspond to the balls.
Therefore,
$$
\left( \! \binom{k}{n} \! \right) = \binom{n + k - 1}{k - 1} = \binom{n + k - 1}{n}.
$$
The only restrictions on the integers $k$ and $n$ are
$$
n \ge 0 \quad \text{and} \quad k \ge 1.
$$
In particular, it doesn't matter which is greater.
I'll solve the case in which the balls are identical. This is not the nicest solution, but it works and I hope it is clear.
Case 1: First two boxes have exactly four balls.
If the first two boxes together contain four balls, then we can use stars and bars twice. The number of ways to place four balls in two boxes is the number of ways to arrange $4$ stars and $1$ bar, or $5$ choose $1$. The number of ways to place the remaining four balls in the other four boxes is the number of was to arrange $4$ stars and $3$ bars, or $7$ choose $3$. This gives
$$\binom{5}{1}\binom{7}{3}$$
Case 2: First two boxes have exactly three balls. With the same reasoning, we get
$$\binom{4}{1}\binom{8}{3}$$
Case 3: First two boxes have exactly two balls.
$$\binom{3}{1} \binom{9}{3}$$
Case 4: First two boxes have exactly one ball.
$$\binom{2}{1}\binom{10}{3}$$
Case 5: First two boxes have no balls.
$$\binom{1}{1}\binom{11}{3}$$
(This makes sense because this is just stars and bars applied to putting $8$ balls into $4$ boxes)
Our total is therefore
$$5\binom{7}{3}+4\binom{8}{3}+3\binom{9}{3}+2\binom{10}{3}+\binom{11}{3}=\boxed{1056}$$
Best Answer
You have reduced the problem to counting strings of $n=8$ stars and $k-1=2$ bars. There are actually 10 places that the bars can go, not 9, so $\binom{10}{2}$ options. This is equal to the $\binom{n+k-1}{n}=\binom{10}{8}$ by symmmetry of Pascal's triangle.