[pzhon]

[ QUOTE ]
An elevator takes on six passengers and stops at ten floors. We can assign two different equiprobable measures for the ways that the passengers are discharged:

(a) we consider the passengers to be distinguishable
(b) we consider them to be indistinguishable

For each case, calculate the probability that all the passengers get off at different floors.

[/ QUOTE ]
Answers in white:

a) <font color="white">There are 10^6 possibilities, and 10*9*8*7*6*5 of them (15.12%) have the passengers get off at different floors.</font>

b) <font color="white">There are (6+10-1)C6 = 15C6 ways to place 6 indistinguishable objects in 10 distinguishable containers. There are 10C6 ways (4.20%) to choose 6 different containers.</font>

I think it is an interesting question to determine which probability should be higher without doing the calculation. Spoiler in white:
<font color="white">Each distribution with undistinguished people corresponds to some number of placements of distinguished people. For example, if 3 people are placed on the first and second floors, there are 6C3=20 ways to distinguish them. The maximum expansion factor, 6!=120, occurs when the people are placed on different floors, so these occur with a greater proportion when you distinguish people.</font>