Let there are $8$ students, $2$ from each of $4$ different schools. They have to be placed in $4$ rooms with $2$ students in each room. What is the probability that students of the same school are in the same room?
My attempt:
The total number of ways in which $8$ students can be assigned in $4$ rooms with groups of $2$ is $$T={8\choose2}4+{6\choose2}3+{4\choose2}2+{2\choose2}1$$
(At first I select two out of the 8 students and place them in one of the 4 rooms then I select 2 from the remaining 6 and place them in the remaining 3 and so on ).
Also, total number of ways in which $4$ groups of $2$ students in each group (students of each school) can be arranged in $4$ rooms is $4!$.
So, $$P=\frac{4!}{T}$$
But this is wrong. Can someone point out why?
Best Answer
You're overcounting because you're multiplying to choose the rooms at each step; instead, the procedure you describe should be "pick the two students for room $1$, pick the students for room $2$, and so on." You'll note that there are multiple ways for you to get the same arrangement here. For example, if the first two students you choose are from school $A$ and put in room $2$, then you would also count this when you count cases where the second two students you choose are from school $A$ and get put in room $2$ (assume all the other assignments are the same). Additionally, you need to multiply when performing multiple steps in order. It should be that $T = \binom{8}{2} \binom{6}{2} \binom{4}{2} \binom{2}{2}$.