How did this shorter method of finding $x^r$ in equation come?

Question

Q)What is coefficient of $x^{30}$ in $$\left(\frac{1-x^{11}}{1-x}\right)^3\left(\frac{1-x^{21}}{1-x}\right)$$

I did this question by expanding $\frac{1}{1-x}$ and multiplying and got a large equation which in the end came to the correct solution of 1111.However, in the book the following steps are provided:

=Coefficient of $x^{30}$ in $(1-x^{11})^3(1-x^{21})(1-x)^{-4}$
=Coefficient of $x^{30}$ in $(1-3x^{11}+3x^{22})(1-x^{21})(1-x)^{-4}$
=Coefficient of $x^{30}$ in $(1-3x^{11}+3x^{22}-x^{21})(1-x)^{-4}$
Here,(according to my understanding) the terms greater than $x^{30} $are excluded as only the terms lower than $x^{30}$ can give the desired result(Expansion of $\frac{1}{1-x}=1+x+x^2+x^3+....$). However, in the next step, it equates the above to:
$$\frac{33!}{30!3!} +(-3)\frac{22!}{19!3!}+(3)\frac{11!}{8!3!}+(-1)\frac{12!}{9!3!}$$which also provides the correct solution. How did this equation come?

Answer 1

$(1-x)^{-n}=1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+...$
Basically, the coefficient of $x^r$ in $(1-x)^{-n} = \binom{n+r-1}{r}$ where n is a natural number

In your case $n=4$
What we want is:
(Coefficient of $x^{30}$ in $(1-x)^{-4}) -(3 \cdot $ Coefficient of $x^{19}$ in $(1-x)^{-4})+(3 \cdot$ Coefficient of $x^8$ in $(1-x)^{-4}) -($ Coefficient of $x^9$ in $(1-x)^{-4}$)
You can take it from here.

Answer 2

When you expand out $(1-x)^{-4}$ you get a sum of the form $$ (1+x+x^2+x^3+\cdots)^4=\sum_a\sum_b\sum_c\sum_d x^{a+b+c+d}\tag1 $$ where the integer indices $a,b,c,d$ run from $0$ to $\infty$. To multiply this on to the four terms in $(1-3x^{11}+3x^{22}-x^{21})$ and collect only powers of $x^{30}$, you'll want to fetch the coefficient of $x^{30}$ in (1) as well as the coefficient of $x^{19}$, and that of $x^8$ and $x^9$.

Think about how many ways you can achieve $x^{30}$ in (1). You need to find four nonnegative integers that sum to $30$. Using a stars-and-bars argument, the answer is ${33\choose 3}$. In general the coefficient on $x^r$ in (1) is ${r+3\choose 3}={r+3\choose r}$.