<s>[INST] How many ways are there for $2$ teams to win a best of $7$ series? Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
Case $3$: $6$ games: Team A wins $4$ games, team B wins $2$ = $\binom{6}{4}\binom{6}{2}-2$...minus $2$ for the possibility of team A winning the first four games; and the middle four (games $2,3,4,5$), in which case there would be no game $6$.
Case $4$: $7$ games: Team A wins $4$ games, team B wins $3$ = $\binom{7}{4}\binom{7}{3}-3$...minus $3$ for the possibility of team A winning the first four games; games $2,3,4,5$; and games $3,4,5,6$.
Total = sum of the $4$ cases multiplied by $2$ since the question is asking for $2$ teams.
Is this correct?
 [/INST] For another approach, if the series ends before the seventh game, extend it to seven games by having the losing team win the rest.  The series will now be four games to three.  There are $2$ ways to choose the losing team, and ${7 \choose 3}$ ways to choose which game the losing team wins.  The total is then $2{7 \choose 3}=70$
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