<s>[INST] Arrange 6 adults and 12 children in 5 rooms, with at least 1 adult in each room Problem: How many ways can you arrange 6 adults and 12 children in 5 rooms of max 4 people such that there is at least 1 adult per room. Every person and room is distinguishable.
My take from the problem is the following: 


*

*There are 2 distinct ways to place 18 people in room of 4: $\{4, 4, 4, 4, 2\}$ and $\{4,4,4,3,3\}$

*For the first sequence, I can start by fixing all the adults in every room in the following manner: $\binom{6}{2} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1} \times \binom{1}{1}$

*Now there are 2 cases: 


*

*The room with 2 adults is the one with 2 people $\binom{12}{0} \times \binom{12}{3} \times \binom{9}{3} \times \binom{6}{3} \times \binom{3}{3}$

*The room where 2 adults is among one that contains 4 people $\binom{12}{2} \times \binom{10}{3} \times \binom{7}{3} \times \binom{4}{3} \times \binom{1}{1}$


*There are $5!$ ways to shuffle the rooms around since they are distinguishable
Adding the 2 cases in (3) and multiplying it with (2) and (4) should in my opinion give me the answer with the sequence $\{4, 4, 4, 4, 2\}$. Is there anything wrong with my reasoning?
 [/INST] You could drive your consideration under 1. even further: you would have the following patterns (using a = adult, c = child):
$A = \{aacc, accc, accc, accc, ac\}$,
$B = \{accc, accc, accc, accc, aa\}$,
$C = \{aacc, accc, accc, acc, acc\}$,
$D = \{accc, accc, accc, aac, acc\}$.
Now let us consider first the rooms to be undistinguishable. Thus the filling pattern can be considered fixed in the above order.
Then we get groupings:
$gA = (\binom{6}{2}\cdot\binom{12}{2})\times(\binom{6-2}{1}\cdot\binom{12-2}{3})\times(\binom{6-3}{1}\cdot\binom{12-5}{3})\times(\binom{6-4}{1}\cdot\binom{12-8}{3})\times(\binom{6-5}{1}\cdot\binom{12-11}{1})$,
$gB = (\binom{6}{1}\cdot\binom{12}{3})\times(\binom{6-1}{1}\cdot\binom{12-3}{3})\times(\binom{6-2}{1}\cdot\binom{12-6}{3})\times(\binom{6-3}{1}\cdot\binom{12-9}{3})\times(\binom{6-4}{2}\cdot\binom{12-12}{0})$,
$gC = (\binom{6}{2}\cdot\binom{12}{2})\times(\binom{6-2}{1}\cdot\binom{12-2}{3})\times(\binom{6-3}{1}\cdot\binom{12-5}{3})\times(\binom{6-4}{1}\cdot\binom{12-8}{2})\times(\binom{6-5}{1}\cdot\binom{12-10}{2})$,
$gD = (\binom{6}{1}\cdot\binom{12}{3})\times(\binom{6-1}{1}\cdot\binom{12-3}{3})\times(\binom{6-2}{1}\cdot\binom{12-6}{3})\times(\binom{6-3}{2}\cdot\binom{12-9}{1})\times(\binom{6-5}{1}\cdot\binom{12-10}{2})$.
Finally you can mix up the room numbers as well, thus you get the sequence counts:
$sA = \frac{5!}{1! \cdot 3! \cdot 1!}$,
$sB = \frac{5!}{4! \cdot 1!}$,
$sC = \frac{5!}{1! \cdot 2! \cdot 2!}$,
$sD = \frac{5!}{3! \cdot 1! \cdot 1!}$.
Thus you get as individual possibilities:
$pA = gA \cdot sA$,
$pB = gB \cdot sB$,
$pC = gC \cdot sC$,
$pD = gD \cdot sD$.  
Or finally as total possibility count:
$P = pA + pB + pC + pD$.
(The actual calculation is kept for you.)
---rk
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