I don't know if it can be done as a single S&B calculation but here are two S&B approaches:
(1) Do S&B for the equation without restriction. Subtract from that another S&B with restriction $x_1 \geq 4$.
(2) Do a separate S&B, omitting $x_1$ from the equation, for each of the four cases: $x_1 = 0,1,2,3$. Then sum the four results.
Example: Take $i=4$ so we have
$$\def\x{x_}\x1+\x2+\x3+x_4 =15\qquad\mbox{with }\x1\leq3$$
(1) S&B without restriction: we have $4-1 = 3$ bars and $15$ stars. #Ways= $\binom{18}{3}$.
S&B with $x_1 \geq 4$: we have $4-1=3$ bars and $11$ stars. #Ways = $\binom{14}{3}$.
TotalWays = $\binom{18}{3} - \binom{14}{3} = 452$.
(2) S&B with $x_1=0$: we have $3-1=2$ bars and $15$ stars. #Ways = $\binom{17}{2}$.
(The equation we have here is: $x_2+x_3+x_4 =15$.)
S&B with $x_1=1$: we have $3-1=2$ bars and $14$ stars. #Ways = $\binom{16}{2}$.
S&B with $x_1=2$: we have $3-1=2$ bars and $13$ stars. #Ways = $\binom{15}{2}$.
S&B with $x_1=3$: we have $3-1=2$ bars and $12$ stars. #Ways = $\binom{14}{2}$.
TotalWays = $\binom{17}{2} + \binom{16}{2} + \binom{15}{2} + \binom{14}{2} = 452$.
Let $10 =a+b+c+d+1,\ i=b+c+d+1,\ j=c+d+1,\ k=d+1$.
where $a,b,c,d$ each represents the 'stars' in every container:
For example in $(∗∗∗|∗|∗∗|∗∗∗)$ $a=3, b=1, c=2, d=3$
Then $\mathrm{the\ number\ of\ permutations} = {9+(4-1)\choose (4-1)} = {12\choose 3} $which apparently is the same as the given one in the textbook.
Then $i$ will be any number from $1$ to $10$, $j$ will be any number from $1$ to $i$, and $k$ will be any number from $1$ to $j$.
*Your formula for stars and bars is not correct.
Source: Brilliant.org
The number of ways to place $n$ indistinguishable balls into $r$ labelled urns is
$$ \binom{n+r-1}{n} = \binom{n+r-1}{r-1}. \ _\square $$
which is not the same as the formula you have given, i.e. ${n + (r-1)}\choose{r}$
Best Answer
If a four-digit number has a digit sum of $9$, then you can consider $9$ stars and $3$ bars, like you say. This means there are $12$ total slots, and choosing the $3$ slots in which the bars go is sufficient for counting the number of ways to arrange them. This is the binomial coefficient $$\binom{12}{3}$$
But we've overcounted (as you noticed), so we need to subtract the number of numbers that we counted that started with zeroes. These are exactly the three-digit numbers (including those with leading zeroes) whose digit sum is nine. We can convert this to $9$ stars and $2$ bars, so our final answer should be
$$\binom{12}{3}-\binom{11}{2}$$
You can also do this problem the way you did it, with $8$ stars and $3$ bars. Why is the answer the same if you do it this way? In other words, why is this true?
$$\binom{12}{3}-\binom{11}{2}=\binom{11}{3}$$
Can you generalize it? After thinking about this you might want to read about Pascal's rule.