|Spoiler Inside: Solution to Puzzle||SelectShow>|
A win for A must end with A. After four games, there is one sequence (AAAA). After five, there are 4!/(3!1!) = 4 (not 5 as AAAAB is not possible). After six, 5!/(3!2!) = 10. After 7, 6!/(3!3!) = 20.
The sequences consist of 3 A’s (so not including the final one) and the remainder B’s with these in every possible order followed by the final A. The fourth A ends the match.
There are 35 sequences. For B’s wins, just reverse A’s and B’s. This gives a total of 70 sequences.