First of all, all permutations and combinations fall into two categories:
1 and a are assigned values respectively.
2.a is assigned another 1 person.
In the case of 1, the other four people can only form teams in pairs. There are three combinations of C (4 4,2)/2 =1* *.
In the case of 2, it can be divided into three sub-categories, that is, Team A with C, D and E, so in each case, the remaining three people also have three combinations, so a * * * has 3x3=9 combinations.
So these five people are divided into three groups as required. A * * has 3+9= 12 combinations, and each combination needs to be arranged according to three schools, which is 3! =6, therefore, we can finally get 12x6=72 distribution methods.