Probability of pushing buttons in elevator 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?
 A: The denominator in this case is $9^3=729$.  This is because it is given that each person's choice of floor is independent, so there is a possibility that more than one of them will want to get off at the same floor.
As noted in comments, the numerator is $7\cdot 3!=42$.  You are correct to note that there are exactly seven ways that the first of the three floors could be chosen.  But you must also consider that A choosing 2, B choosing 3, and C choosing 4 is a different event than A choosing 3, B choosing 4, and C choosing 2 even though the floors visited are the same in both cases.
That gives us a total probability of $42\over729$.
A: The sample space has to include events such as $(2,2,3)$ and even $(5,5,5)$ since it was nowhere said that the buttons pressed are different. In fact, the fact that they are independent implies that they can be equal in this case.
So the sample space has naturally $9\cdot9\cdot9$ elements, and it's more natural to count ordered triples rather than unordered, because not all triples have the same symmetry.
A: 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
$$3!\cdot 7\left(\frac{1}{9}\right)^3 = \frac{42}{729}.$$
