Counting problem about housing suppose there are 3 rooms in a dormitory: one single, one double, and one for four students. How many ways are there to house 7 students in these rooms?
The answer is 7$\times$(5+4+3+2+1), but I don't understand why, can anyone explain this to me?
 A: The reasoning behind the answer $7\times(5+4+3+2+1)$ must be something like this:
First we choose which student to put in the single room. There are $7$ possible choices for this.
Then we choose the older of the two students in the double room. The youngest of 6 students left by now cannot possibly be the older in the double room, but the 5 others are fair game. That gives of $5$ possible choices in each phase. But depending on which student we chose to be the older, there's a different number of possible choices for the younger student in the double room, so instead of just multiplying something by $5$ we need to add the number of choice-of-younger-student for each choice-of-older-student, giving $5+4+3+2+1$ ways to fill the double room.
Once the single and double room are occupied, the 4 remaining students go in the last room, and there's only one way to do that.
A: One way to arrange the students is to first choose the students who will inhabit the four-student room in $\binom{7}{4}$ ways, then choose two of the remaining students to inhabit the two-student room in $\binom{3}{2}$ ways. The remaining student will be forced into the single-student room. Since we made one choice followed by another choice, the total number of outcomes is $\binom{7}{4} \binom{3}{2} = 105$.
