Question: There is a ball drawer.  

Seven color balls will be drawn with the same probability ($1/7$).  

(black, blue, green, yellow, white, pink, orange)

If Anson attempts $9$ times,  

what is the probability that he gets all $7$ different color balls?

My work:

I separate the answer to $3$ ways.

1. $7$ attempts -> done (get $7$ colors)
2. $8$ attempts -> done (get $7$ colors)
3. $9$ attempts -> done (get $7$ colors)

Therefore, my answer is $$\frac{9C7 + 8C7 + 7C7}{7^7 \cdot (7!)}$$
However, I don't know it is correct or not.

Answer: 
The statement is equivalent to
$$\DeclareMathOperator{\arccot}{arccot}
\arccot7+\arccot8=\arccot3-\arccot18
$$
The right-hand side is positive because the arccotangent is decreasing, so it is in the interval $(0,\pi/2)$ and we have to ensure that the same holds for the left-hand side, but this is easy, because $7>1$ and $8>1$, hence
$$
\arccot7+\arccot8<2\arccot1=\frac{\pi}{2}
$$
The two sides are equal if and only if their cotangents are.

Since
$$
\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}
\qquad
\cot(\alpha-\beta)=\frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}
$$
this amounts to proving that
$$
\frac{7\cdot8-1}{7+8}=\frac{3\cdot18+1}{18-3}
$$
I can't resist, even though your request for a critique makes this response somewhat off-topic. Let the 3 LHS angles be denoted as $a,b,c$, and use the formula

$$\cot(a + b) = \frac{[\cot(a)\cot(b)] - 1}{\cot(a) + \cot(b)}.$$

This gives $\cot(a + b) = \frac{55}{15} = \frac{11}{3}.$

Therefore,
$$\cot[(a+b) + c] = \frac{\left[\left(\frac{11}{3}\right)\left(18\right)\right] - 1}{\frac{11}{3} + 18} = 3.$$

Further, since each of $a,b,c$ are in the first quadrant, as is $(a + b + c)$, you have that $(a + b + c) = \text{Arccot}(3).$