(1) All three rooms need to be checked in. According to the topic, the single room must be one adult (in three different ways), and the remaining two adults and two children live in triple rooms and double rooms.

A. adults +2 distribution methods between children and adults (if BC is left, there are two kinds of companionship methods, namely B+ab and C; C+ab and b), that is, adults +2 children must live in three rooms.

B. Distribution modes of adults+children and adults+children (these are four modes B+A; b+ b； c+a； C+b), that is, a triple room and a double room, one adult and one child.

The total * * * distribution mode is 3*(2+4)= 18.

There are 18 different ways to arrange accommodation.

(2) The known conditions are changed to the problem of allocating five people to live in triple rooms and two double rooms. Every child must bring at least one adult.

A. Adults +2 children and two adults living together (these are three ways for ABC to accompany two children respectively) need two rooms, that is, two distribution methods (three rooms for adults +2 children and two double rooms for adults) 3*2=6.

B. Adults +2 children live separately from two adults (this is three ways for ABC to accompany two children respectively), and all three rooms are occupied. There are two distribution methods (three adult rooms+two children, a double room and one room for two other adults) 3*2=6.

C adult+1 children and adults+1 children and adults (this is 12 ways 3 * 4; At this time, there are three situations (single adults live in 3 rooms and single adults live in 2 rooms) 12*3=36.

D. Adults+1 children and two adults+1 children (this is six ways for adults to choose a pair of children, that is, 3*2). There are only two situations (three people must give two adults+1 children) 6*2= 12.

6+6+36+ 12=60

To sum up, the total is 60 cases.