An urn contains n red balls, n white, n black. What is the probability of not getting all colors? I have this problem which I have been struggling with for a while now

An urn contains n red balls, n white balls and n black balls. You
  draw k balls at random without replacement ($k\leqq n$). Find an
  expression for the probability that you do not get all colors.

I tried to solve this in the following way:
I note that it should be logical to think P(not getting all colours in k draws) = P(getting exactly one color only in k draws OR get exactly two different colors only in k draws).
Therefore, I choose the events $A_1=\{$get one color k times$\}$ and $A_2=\{$get two colors k times$\}$
Therefore, we seek $P(A_1 \cup A_2)$.
Clearly, $A_1 = \frac{{n\choose k}}{3n\choose k}$ and  $A_2 = \frac{{2n\choose k}}{3n\choose k}$. And because the both events are disjoint, we simply get 
$P(A_1 \cup A_2) = P(A_1) + P(A_2) = \frac{1}{3n\choose k}\big({n\choose k}+{2n\choose k}\big)$.
According to my textbook, however, the answer should be
$\frac{3}{3n\choose k}\big({2n\choose k}-{n\choose k}\big)$.
What am I doing wrong? Are the answers equivalent because of some mystical binomial-identity?
 A: You have to multiply by 3, because there are three ways to select the colors. Furthermore, when considering all ways to pick two colors, you also include the scenario where you select only one color. These must be subtracted, hence you get:
$$\frac{{3 \choose 2}\left({2n \choose k}-2{n \choose k}\right)+{3 \choose 1}{n \choose k}}{3n \choose k} = \frac{3({2n \choose k} - {n \choose k})}{3n \choose k}$$
A: Let us take the colors blue, red and green and let $B,R,G$ stand for the events that no blue, red, green balls are selected respectively.
Then you are looking for $P(B\cup R\cup G)$ and with inclusion/exclusion and symmetry we find:$$P(B\cup R\cup G)=3P(B)-3P(B\cap R)=3\left[\frac{\binom{2n}{k}-\binom{n}{k}}{\binom{3n}{k}}\right]$$
A: The     reader      may     want     to     consult      this     MSE
link  by   way  of
enrichment    where    a    much    more    general    question    was
answered. Substituting the parameters  from the present question value
by  value into  the  formula  that was  proved  there  yields for  the
complementary probability of seeing all  three colors the form (in the
following we have $n\ge 1$)
$${3n\choose k}^{-1} {3\choose 3}
\sum_{p=0}^3 {3\choose p} (-1)^{3-p} {pn\choose k}
\\ = {3n\choose k}^{-1}
\times
\left({3n\choose k} - 3{2n\choose k}
+ 3{n\choose k} - {0\choose k}\right).$$
This gives  for $k=0$ (no balls  drawn) the probability of  seeing all
three colors
$$1\times ( 1-3+3-1 ) = 0.$$
For $1\le k\le 3n$ we obtain
$${3n\choose k}^{-1}
\times
\left({3n\choose k} - 3{2n\choose k} + 3{n\choose k}\right).$$
The complementary probability then becomes
$$1 - {3n\choose k}^{-1}
\times
\left({3n\choose k} - 3{2n\choose k} + 3{n\choose k}\right)
\\ = {3n\choose k}^{-1}
\left( 3{2n\choose k} - 3{n\choose k} \right).$$
This confirms the two answers that were first to appear.
