TITLE: Probability of pushing buttons in elevator
QUESTION [7 upvotes]: I am considering the following problem from Introduction to Probability by Blitzstein and Hwang (Exercise 23 of Chapter 1).

Three people get into an empty elevator at the first floor of a
  building that has 10 floors. Each presses the button for their desired
  floor. Assume that are equally likely to want to get to floors 2
  through 10 (independently of each other). What is the probability that
  the buttons for 3 consecutive floors are pressed.

My thoughts
There are 3 people A, B and C in the elevator. If A pushes a, B pushes b and C pushes c (a triple (a,b,c) where b=a+1 and c=b+1), then the number of total triples that can be formed from a sample of $9$ elements is $\binom93=84$. I do not see why the sample space should be $ 9\cdot9\cdot9$ since we are only interested in the triples and more specifically we are interested in the ordered triples. 
The favorable outcomes are (2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9), (8,9,10), i.e. 7 in total. 
The probability that the buttons for three consecutive floors are pressed is therefore $\dfrac{7}{84}
$.
Is the above correct?

REPLY [4 votes]: I think the others have offered good explanations of the sample space. I'd like to offer a different approach that might be a little more intuitive. 
I like to think of these elevator problems in terms of throwing dice. In other words, you throw a 9-sided die 3 times. What is the probability that the throws form a consecutive trio?
Well, first consider the probability of $(2,3,4)$, 
$$P(2,3,4) = \left(\frac{1}{9}\right)^3,$$
where it's the product by independence.
But it didn't have to be in that order. So we multiply by a factor of $3!$. 
Next, it didn't have to be those three. As you noted, there are 7 triplets we could choose from. Therefore, the probability is
$$3!\cdot 7\left(\frac{1}{9}\right)^3 = \frac{42}{729}.$$