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).