Basic question about combinatorics and probability For the following question:

Six race cars, each a different color, race each other around a track.
  If the order in which they finish is random, what is the probability
  that the turquoise car finishes before both the teal and the
  aquamarine cars?

The answer in the textbook is 1/6. However, I do not understand how to arrive to that solution (I am getting 1/18). Here was my thinking:
There are 4 possibilities for where the turquoise car could be:
_ _ _ _ _ T
The two cars (teal and aquamarine) have to be behind T, so there are 5 * 4 = 20 ways to choose the positions of the 2 cars.
_ _ _ _ T _
4 * 3 = 12 ways
_ _ _ T _ _
3 * 2 = 6 ways
_ _ T _ _ _
2 * 1 = 2 ways
In total, that means there are 20 + 12 + 6 + 2 = 40 ways. The probability of each way is (1/6! = 1/720). 40 * 1/720 = 40/720 = 1/18. 
 A: Because one of the three cars in question must lead the other two and they are presumed to have equal chances, the answer must be $1/3.$
We can't determine why the textbook is wrong, but your analysis is clear enough that we can indicate how to fix it.  The problem is that in counting the possibilities you have not accounted for the positions of the other three cars, but in dividing that count by $6!$ you do account for their positions.
Given any of your configurations, such as _ _ _ _ T _, and given the positions of the other two cars, there are always three slots remaining in which the other three cars must fall: that gives $3!=6$ possibilities in each case.  Thus, the correct calculation using your method must be
$$\eqalign{&\Pr(\text{Turquoise precedes Teal and Aquamarine})\\ &= \frac{5\times 4 \times 3!}{6!} + \frac{4 \times 3 \times 3!}{6!} + \frac{3\times 2 \times 3!}{6!} + \frac{2\times 1\times 3!}{6!} = \frac{1}{3}.}$$

If any of this gives you a queasy feeling--and that's not unusual with probability problems or sloppy textbooks--often a quick simulation can be a guide.  Here's a simulation of a million race outcomes in R.  The positions of the cars are determined by identical, independent random values which have (essentially) zero chance of being tied.  On its ultimate line the code calculates the fraction of these races in which the Turquoise value exceeds the values of both Teal and Aquamarine:

n <- 1e6
cars <- c("Turqoise", "Teal", "Aquamarine", "Chartreuse", "Ecru", "Heliotrope")
set.seed(17)
x <- matrix(runif(length(cars)*n), ncol=n, dimnames=list(cars))
mean(x["Turqoise",] > x["Teal",] & x["Turqoise",] > x["Aquamarine",])


[1] 0.332709

That should be convincing evidence of the $1/3$ answer.
