Multinomial Probability Question A lift starts with 5 passengers and stops at 8 floors. Find the probability that no two or more passengers leave at the same floor. Assume that all arrangements of discharging passengers have the same probability.
Answer:-  The probability that no two or more passengers leave at the same floor is $\frac{6720}{32768}.$The various arrangements of discharging the passengers to any one floor may be denoted by symbols like (3,1,1) to be interpreted as the event that three passengers leave together at a certain floor, one passenger at another floor and finally the last one passenger at still another floor. So there are seven possible arrangements ranging from (5) to (1,1,1,1,1). Now calculate the total probability by adding probability for each arrangement. E.g. In how many ways all the five passengers gets off at any one floor. The answer is in 
  $\binom{8}{1}$ ways. In this way,you can arrive at my final answer $\frac{6720}{32768}$
 A: I'm not able to comment and this is my first time answering a self-study so let me know if changes are needed.
My interpretation of this problem is "all arrangements of discharging passengers" refers to all possible outcomes i.e. Passengers A, B, C, D, E getting off on (5th floor, 5th floor, 2nd floor, 3rd floor, 7th floor) or  (3rd, 4th, 5th, 6th, 7th) or (1st, 1st, 1st, 1st, 1st), etc. 
The problem is looking for P(no more than 1 passenger per floor) = (outcomes that meet the criteria / all outcomes). The outcomes that meet the criteria is a without replacement combinatorics problem. 


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*You can not replace the passengers but you can replace the floors in the non-criteria meeting instances of more than one passenger exiting on a given floor e.g. (1st, 1st, 1st, 1st, 1st). 

*My interpretation is everyone will get off the elevator. I think the inclusion of the no passenger gets off any floor is not in the spirit of the question.

*Outcomes that meet that criteria $=$ instances where all the floors selected will be different.
Another hint/correction: You can select the 5 destination floors from the 8 available floors more than ${8 \choose 5}$ ways. 
