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
$$
\frac{7\cdot8-1}{7+8}=\frac{3\cdot18+1}{18-3}
$$
${tan^{-1}(1/7)+tan^{-1}(1/8)+tan^{-1}(1/18)}$

* ${tan^{-1} \left[ \frac{ (1/7)+(1/8) } { \left( 1 \right) - (1/7)(1/8) } \right] + tan^{-1}(1/18) }$
* = ${tan^{-1} \frac {3}{11} + tan^{-1} \frac{1}{18}}$
* =${tan^{-1} \left[ \frac {(3/11)+(1/18)} {\left (1 \right) - (3/11)(1/18)} \right] }$
* = ${tan^{-1} \frac{1}{3} }$ or ${cot^{-1} {3}}$