What is the total number of ways for 5 people to occupy 4 rooms

Question

What is the total number of ways for 5 people to occupy 4 rooms where one of the rooms is occupied by two people and each of the remaining 3 rooms by one person? Please write down intermediate steps.

Answer 1

Like many counting problems, this one can be tackled by breaking it into independent steps.

First choose the room that will hold two people. There are four ways to do this since any of the four rooms can be chosen. Importantly, this choice doesn't affect any of our future choices.

Next, choose which two of the five will be together in the room you chose. There are "five choose two", or ten ways to do this. Explicitly, this is 5!/(3!*2!).

Once this choice is made, the only remaining choice is which of the three remaining people goes in which of the three remaining rooms. If we arbitrarily label the rooms 1, 2 and 3, then this amounts to ordering the three people. The number of ways to order three people is 3! or six.

Each of these steps is independent, and you should check that we're not overcounting or undercounting. That is, there's no possibility missed by making the choices this way, and there's no way that two different ways of making these choices could end in the same configuration of people and rooms.

So just multiply the choices at each step. We get 4*10*6 = 240 ways.

Lastly, if we reflect on our answer rather than blazing on to the next problem, we can often find some further understanding. For instance, there's a 3! in the denominator of the second step, and in the numerator of the third step. The fact that these cancel suggests a simpler way of looking at it. There are four ways of selecting the special room, and then there are 5!/2! ways of choosing, in order, the three people who won't be put in that special room. The reason order matters is because the rooms are distinct.