Question 917936
i. Casework on # games played (we only count # outcomes with 4 W's)

4 games: Only WWWW, 1 outcome
5 games: Any rearrangement of WWWWL with the last game W, 4C1 = 4 outcomes
6 games: Any rearrangement of WWWWLL, with the last game W, 5C2 = 10 outcomes
7 games: Any rearrangement of WWWWLLL, with the last game W, 6C3 = 20 outcomes


1+4+10+20 = 35. Multiply by 2 to account for outcomes with 4 L's --> 70 different outcomes

ii. 71 by Pigeonhole


iii. Equivalent to the number of 4-tuples whose sum is 17 and each element is at least 1 (since it is impossible to win with 0 runs). This is equal to (13+3)C3 = 16C3 = 560 (look up stars and bars if you're not familiar)