Question 1100178
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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 {{{7^7}}} such words.
Correspondingly, there are {{{7^7}}} 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  

{{{(7*6*5*4*3*2*1)/7^7}}} = {{{7!/7^7}}} = {{{5040/823543}}} = 0.00612 = 0.612% (approximately).
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Solved.