We have two rooms painted white and one painted grey. In how many ways can we arrange 90 I have a combinatorics problem, I tried to solve it myself, but I don't know if it's correct.
We have two rooms painted white and one painted grey.
In how many ways can we arrange 90 people (30 in each room) so that:

*

*it is important for us who is with whom in the room and whether the person is in a white or gray room

*both white rooms are indistinguishable for us

What I tried to do:

*

*First select 30 people out of 90 (grey room) i.e.: $\binom{90}{30}$

*then draw a group of people to the white room: $\binom{60}{30} \times \binom{30}{30}$
Do divide by 3! (because when I choose 30 people for the gray room, I automatically choose 30 for the white one and 30 for the white one2?) expression : $\binom{90}{30} \times \binom{60}{30} \times \binom{60}{30}$
Please explain how to go about it and what the fact that white rooms are indistinguishable brings to the task
 A: Since only the two white rooms are indistinguishable, you should be dividing by $2$ rather than $3$.
If each of the rooms were distinguishable, we could place $30$ people in each room in
$$\binom{90}{30}\binom{60}{30}\binom{30}{30}$$
ways.  Since the white rooms are indistinguishable, we could switch the people in the two white rooms without changing which people are together.  Thus, we have counted each such arrangement twice.  Hence, there are
$$\frac{1}{2}\binom{90}{30}\binom{60}{30}\binom{30}{30}$$
distinguishable ways to distribute the $90$ people to one grey room and two indistinguishable white rooms so that each room receives $30$ people.
Alternatively, suppose that we have placed $30$ of the $90$ people in the grey room and that Amanda is one of the remaining $60$ people.  The two groups in the white rooms are distinguished by whether or not Amanda is a member of the group.  There are $\binom{59}{29}$ ways to select the other $29$ people who will be placed in the same white room as Amanda.  The remaining $30$ people must be placed in the remaining white room.  Hence, there are
$$\binom{90}{30}\binom{59}{29}$$
distinguishable ways to distribute the $90$ people to one grey room and two indistinguishable white rooms so that each room receives $30$ people.
