Probability Counting Problem Confusion Im having difficulty understanding how to do the following problem using Combinations.
Q: An urn contains 2 red balls, 4 blue balls and 6 green balls. If three balls are drawn at random one at a time, without replacement, what is the probability that the first and last are the same colour?
The book answer solves this by considering that the third choice of a ball does not affect the experiment and treats this experiment as of choosing two balls instead of three and thus solving the problem as: $\dfrac{{2\choose2}+{4\choose2}+{6\choose2}}{12\choose2}=\dfrac{1}{3}$ 
Now I was trying to solve this question without disregarding the third ball. My attempt to solve this was: $\dfrac{{10\choose1}[{2\choose2}+{4\choose2}+{6\choose2}]}{12\choose3}=1$ which is obv wrong. My motivation was the following $P(RR \cup BB \cup GG)=P(RR)+P(BB)+P(GG)$ where for example $P(BB)=\dfrac{{4\choose2}{10\choose1}}{{12\choose3}}$ since I have to fill two spots with blue balls and I have 4 choose 2 ways of doing that, however for each of those outcomes I have to fill the last spot with an arbitrary ball, now since I chose 2 blue balls, I am left with 10 balls In the urn and thus 10 choose 1 options to fill the last spot, so therefore the number of ways I can choose 2 blue balls and an arbitrary third ball is 4 choose 2 multiplied by 10 choose 1 ways. Now I divide that by the total number of ways of choosing 3 balls from 12. I carry the same analogy for the Red and Green situations. 
I am just confused where Im making a mistake in my thinking and how to solve this qs without disregarding the last third ball.
Thank You in advance
 A: The book's answer can ignore order. Your answer cannot. You need specifically for the first and last ball to be the same. For the book's answer, they are only looking at the first and last ball, and they do not care which order they pull them so long as they are the same color. Once you add in the third ball, order matters. So, you have RAR, BAB, GAG with A being any color. Choose the first ball, then the last ball, then the middle ball, but order matters, so you are assigning balls to positions, and therefore cannot use combinations. 
$$\require{cancel} \begin{align*}\dfrac{2\cdot 1\cdot 10 + 4\cdot 3\cdot 10 + 6\cdot 5\cdot 10}{12\cdot 11\cdot 10} & = \dfrac{2\cdot 1\cdot 10 + 4\cdot 3\cdot 10 + 6\cdot 5\cdot 10}{12\cdot 11\cdot 10}\cdot \dfrac{\tfrac{1}{2\cdot 10}}{\tfrac{1}{2\cdot 10}} \\ & = \dfrac{2\cdot 1\cdot \cancel{10}\cdot \dfrac{1}{2\cdot \cancel{10}} + 4\cdot 3\cdot \cancel{10}\cdot \dfrac{1}{2\cdot \cancel{10}} + 6\cdot 5\cdot \cancel{10}\cdot \dfrac{1}{2\cdot \cancel{10}}}{12\cdot 11\cdot \cancel{10}\cdot \dfrac{1}{2\cdot \cancel{10}}} \\ & = \dfrac{\dbinom{2}{2} + \dbinom{4}{2} + \dbinom{6}{2}}{\dbinom{12}{2}}\end{align*}$$
So these two methods give the same answer, which is apparent after some algebra to show how they are the same.
