Mathematics Asked by user747916 on December 25, 2020

An elevator in a building can stop at floors $1$ through $6$. Four people get on the elevator at floor $1$ and each person picks a floor at random (from floors $2$ to $6$) and gets off. What is the probability that nobody gets off at odd-numbered floors? What is the probability that half the riders get off at the same floor and the other half get off at a different floor?

For this question, I’m unsure whether to assume the people are indistinguishable (i.e. if person 1 is on floor 2 and persons $2-4$ are on floor $3$, then it’s considered the same as if person $2$ is on floor $2$ and persons $1,3,4,5$ are on floor $3$). If they are not indistinguishable, the problem gets more challenging. For these kinds of problems, does one usually assume objects are indistinguishable if that is not specified?

However, if they are indistinguishable, then the problem becomes much easier. There are $4$ stars representing the people and $4$ bars to separate the $5$ floors, for a total of ${8choose 4} = 70$ possibilities. The cases where nobody gets off at odd floors can be represented by $a_1||a_2||a_3,$ of which there are ${6choose 2} = 15$ ways to choose tuples $(a_1′, a_2′,a_3′)$, where $a_i’$ is the number of stars $a_i$ has (this is just stars-and-bars with $4$ stars and $2$ bars). So the probability would be $15/70 = boxed{3/14}.$ Similarly, for the other question, there would be ${5choose 2} = 10$ ways to choose the two different floors. Since the people are indistinguishable, these are the only distinguishable events, for a probability of $10/70 = boxed{1/7}.$

I’m quite sure that my solution for the indistinguishable case is correct, but is it? How do I deal with the case where the people are distinguishable? That’s like stars-and-bars except with distinguishable stars. Can this case be generalized well?

Edit: I’m taking a guess here, but for the distinguishable case, I think there are $5$ ways to select $1$ floor for all four people, $10$ ways to select two different floors, and for each different floor, we may have $6$ possibilities where there are $2$ people per floor, or $2times 4$ possibilities where $1$ person is on one floor and the rest are on the other (multiply by $2$ as we can swap). There are also $10$ ways to select three floors, and for each way, there are $3$ ways to select which floor has $2$ people, $6$ ways to choose those two people, and two ways to position the remaining two. Finally, there are $5$ ways to select four floors and $24$ ways to position those people on those floors. This gives a total of (I’m really not sure) $5 + 10 cdot (6 + 2cdot 4) + 10 cdot (3 cdot 6 cdot 2) + 5 cdot 24 = 625$ ways???

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