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.

## Comments

It is the number of non-negative integral solutions of x_1+x_2+x_3+x_4+x_5=5

i.e. 9C3 will be the answer.

For the second part,

each person has 5 ways to choose his floor

Therefore 5 persons can arrange in 5^5 ways

It is the number of non-negative integral solutions of x_1+x_2+x_3+x_4+x_5=5

i.e. 9C4 will be the answer.

For the second part,

each person has 5 ways to choose his floor

Therefore 5 persons can arrange in 5^5 ways