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

$(1-x)^{-n}=1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+...$
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.
