[Math] Five people get on the elevator that stops at five floors. In how many ways they can get off

combinatorics

Five people get on the elevator that stops at five floors. In how many ways they can get off? For example, one person gets off at the 1st floor, two will get off at the third, and the remaining two at the fifth floor. In how many ways they can get off at different floors? Now, consider that people in elevator have names, say A,B,C,D and E., assuming that, for example, the case A on the 1st floor is different from the case B on the 1st floor. Answer the previous questions with this assumption.

I was told that there are 4 questions, but I'm not really sure about it.

People don't have names and get off
People don't have names and get off at different floors
People have names and get off
People have names and get off at different floors.

Do the names matter?

I think I can answer one question out of 4 – the number of all outcomes should be $5^5=3125$. I am confused about the rest though

Best Answer

This has been answered in other answers: ${9\choose4}$.
Since there are $5$ people and $5$ floors there is just $1$ way for them to get off at different floors: $1$ per floor.
You already found $5^5=3125$.
At each floor exactly one of $A$, $\ldots$, $E$ gets off. This can be done in $5\cdot4\cdot3\cdot2=5!=120$ ways.

Related Solutions

[Math] In how many ways can people get out of the elevator

The numbers are small, so we can do it by cases. Maybe (Case 1) $4$ people get off at one floor, with the rest $1$ each. Or maybe (Case 2) it is $3$ on one floor, $2$ another, and the rest $1$ each. Or maybe (Case 3) it is $2$. $2$, $2$ with the rest $1$ each.

I will do Case 1, and Case 3 because it is slightly tricky. You can do Case 2.

Case 1: The floor at which the $4$ people get off can be chosen in $\binom{5}{1}$ ways. For each such choice, the actual people who get off there can be chosen in $\binom{8}{4}$ ways. Once this is done, there are $4$ floors and $4$ people. That gives $4!$ arrangements. So there are $$\binom{5}{1}\binom{8}{4}4!$$ Case 1 possibilities.

Case 3: We do this one because it is tricky, one can get the wrong answer. The special floors at which the pairs get off can be chosen in $\binom{5}{3}$ ways. Once these have been chosen, the $2$ people who get off at the lowest special floor can be chosen in $\binom{8}{2}$ ways and then the $2$ people who get off at the next special floor can be chosen in $\binom{6}{2}$ ways, and then the $2$ people who get off at the highest chosen floor can be selected in $\binom{4}{2}$ ways. Finally, the two singles can be assigned to the remaining floors in $2!$ ways. Multiply.

[Math] An elevator starts with 10 people, how many ways can all the people… Cases for each floor

look up "stars and bars" - you are placing 10 balls in 7 boxes ( unless you really want people to be able to get out on the first floor, in which case you have 8 boxes )

Most likely the question intends that you think of "a way" as 6 people getting out on the 4th floor and 4 people getting out on the 5th floor and is not making the distinction about exactly which people are getting out on every floor.

If we go with 7 floors we will have 8 bars separating the floors, and we must arrange 10 stars representing people getting out. so the example above would be represented as ...

floor #  2 3     4       5   6 7 8
        | | | ****** | **** | | | |

You must keep the first and last bars in the first and last positions, but other than that, any arrangement of 10 stars and 6 bars corresponds to a possible scenario the number of arrangements is

$$ \frac{16!}{10! 6!} = \binom{16}{6} $$

if you really wanted to let people out on the first floor the answer would be $\binom{17}{7}$

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[Math] In how many ways can people get out of the elevator

[Math] An elevator starts with 10 people, how many ways can all the people… Cases for each floor

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