Generally speaking, remove A running first and B running fourth, and then add A running first and B running fourth.

A(4，6)-2*A(3，5)+A(2，4)

=360- 120+ 12

=252 methods

I am puzzled by the method given in the answer, that is, when both parties participate in the competition, it is the method of C (2,4)+C (2,4) * C (1,2) * C (1, 2) * A (2,2).

First kind

C(2，4)*A(3，3)

It means to choose two people from four people other than A, B and C, C(2, 4).

Then A runs the fourth bar. At this time, B and two other people are infinite, which is A (3,3).

the second type

C (2,4) * c (1,2) * C (1, 2) * a (2,2) means to choose two people from four people except Party A, and c (2,4).

Then A doesn't run the fourth bar, but he can't run 1 bar either, so there is C( 1, 2).

At this time, there are still three bars that nobody runs, but B can't run the fourth bar, so there is also C( 1, 2).

The other two are infinite, and they are A (2,2).