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[LetYouDown]

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.

[bobman0330]

Distinguishable case is not hard:
<font color="white">Assign each passenger, A-F, a floor they will get off on.
P(A's floor is unique) = (9/10)^5
P(B's floor is unique|A's floor is unique) = (8/9)^4
P(C's floor is unique|A and B are unique) = (7/8)^3
...
P(All are unique = (after cancellation) (9*8*7*6*5)/10^5 = 32.240% (done by hand, could be wrong) </font>

The indistinguishable case is harder:
<font color="white">At each floor, every passenger has a 1/(# of remaining floors, including the current one) of getting off.

The way I approached this problem is to start at the last floor. At this point, if there are 0 or 1 passengers left and all prior moves have been "legal," the condition has been satisfied. If there are 2 or more, or any illegal moves, it has not. Now, the key is to go through all legal moves that can result in either of these states and add up their probabilities. (Notation: [x, y] = P(either of the winning states will be arrived at legally with x passengers on floor y). The last floor is floor 1, the first is floor 10.)

So, [0,1-4]=[1,1-5]=1
In general, [x,y] = (x*(1/y)((y-1)/y)^(x-1)*[x-1, y-1] + ((y-1)/y)^x*[x, y-1].
The second term will often = 0. E.g., the second term of [6,6] = 0, because [6,5] cannot win.
Iterate away to find [6,10].
</font>

[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>