The coefficient of $x^{100}$ in the expansion of $(1-x)^{-3}$. Please solve it and tell me the technique so that I can solve it in examination in multiple choice questions.

We begin by noting that $\frac{1}{1-x}=(1-x)^{-1}=1+x+x^2+x^3+x^4+\dots$
We see then that $(1-x)^{-3}=(1+x+x^2+x^3+\dots)(1+x+x^2+x^3+\dots)(1+x+x^2+x^3+\dots)$
Approaching from a combinatorial perspective, we recognize that to have arrived at the term $x^n$, we will have needed to selected a term from the first parenthesis, the second parenthesis, and the third such that their exponents add up to $n$.  We further recognize that each such way of doing this will contribute one to the overall coefficient and there are no other ways to arrive at that term.
The question then is related to:

How many non-negative integer solutions are there to the system $$\begin{cases}x_1+x_2+x_3=n\\0\leq x_i\end{cases}$$

From earlier example, we know the answer to be $\binom{n+3-1}{3-1}$ via stars and bars.
For the specific example of $n=100$, this would imply the coefficient to be $\binom{102}{2}=5151$
