Combinatoric Homework Problem

[PairTheBoard]

chiachu:"An elevator with 9 passengers stops at 5 different floors. If we are only interested in the passengers who get off together, how many possible distributions are there?"

Especially: "we are only interested in the passengers who get off together"

I don't see that anyone has correctly read the problem. As stated, the different combinations are determined by the subgroupings of identifiable passengers.

For example, suppose the passengers are A,B,C,D,E,F,G,H,I. Suppose A,C,E,G,I get of on Floor 2 and the rest, B,D,F,H get off on Floor 4. That's One distinct combination we're "interested in". Now suppose B,D,F,H get off on Floor 2 and the other group A,C,E,G,I get off on Floor 4. That's NOT another distinct combination because "we are only interested in the passengers who get off together" not on which floor they do so. HOWEVER, if A,B,C,D,E get off on Floor 2 and F,G,H,I get off on Floor 4 that IS another distinct combination because different passengers have gotten off together.

I would assume that passengers may get off on All Five Floors. There is nothing in the problem which specifically disallows departure on one of the floors so I would not try to conclude such a restriction. Nothing is said about when the 9 passengers got on the Elevator nor whether some of the 9 might want to get off on Floor 1 of the 5. If floor 1 is the Ground Floor it's possible some passengers get on to make the 9 and some then get off from their trip down. The KISS version is that 9 passengers are on the elevator going up where 5 stops are made. In either case I would assume that after visiting all 5 stops that all passengers have reached their floor and will have departed.

I believe this is the correct reading as chiachu states the problem. If he has paraphrased the problem incorrectly rather than giving us an exact quote for it then all I can say is, may the Lord have Mercy on his Soul.

I leave it to those who feel like the exercise to give a correct solution to the correct reading of the problem.

PairTheBoard

[chiachu]

wow i didnt even know this thread was still going. [img]/images/graemlins/grin.gif[/img]

i quoted the problem exactly, but it is written terribly. Especially that one line that 'pairtheboard' quoted. Our profesor actually told us not to do the problem a few days ago, but im still curious to the solution.

After reading 'pairtheboard' take on the problem, i think he is correct in interpreting the question.

[PairTheBoard]

I'm afraid I can't see a slick way to do it ciachu. Grinding it out looks to be a real chore to me.

There's 1 way they can get off on 1 floor.

The number of ways they can get off on 2 floors is
C(9,8) + C(9,7) + C(9,6) + C(9,5) accounting for when they get off in groups of (8,1),(7,2),(6,3),(5,4).

Getting off on three floors can be done in groups of
(7,1,1) (6,1,2) (5,3,1) (5,2,2) (4,4,1) (4,3,2) (3,3,3)

where care must be taken in counting the ways eg. for (3,3,3) as C(9/3)C(6/3)/2 to eliminate double counting with order.

The rest can be ground out similiarly. If there's a slick way to do it I'd like to see it. Hell chiachu, you're the one studying combinatorics. You come up with it.

PairTheBoard