.
Imagine that all and each floors from 2 to 8 are marked by the first 7 letters of English alphabet:
2 3 4 5 6 7 8
A B C D E F G
Then the space of all possible events is the set of all 7-letter words comprising of these letters.
Repetitions of letters in these words are allowed.
It is easy to calculate the number of all such 7-letter words.
Any of 7 letter can stay in the 1-st position. This gives 7 opportunities.
Any of 7 letter can stay in the 2-nd position. This gives 7 opportunities.
And so on . . .
In all, there are
such words.
Correspondingly, there are
elements in the space of events, in all.
Now, the winning events are those 7-letter words what have no repetitions.
The number of such words is exactly 7*6*5*4*3*2*1 = 7! = 5040.
Therefore, the probability under the question is equal to
=
=
= 0.00612 = 0.612% (approximately).
Solved.