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?