Two chess players, A and B, are going to play 7 games. There are three possible outcomes for each game, A wins, A loses, or Tie

Addtionally, a win is worth 1 point, draw 0.5 points and loss 0 points. (a) How many possibilities for the games are there such that player A ends up with 3 wins, 2 draws and 2 losses? (b) How many possible outcomes for the games are there such that A ends with four points and B ends with 3 points. (c) Now, assume the first player to 4 points wins and no further matches are played. How many outcomes are there such that the match lasts 7 games and A wins with a score of 4 to 3?

(a) There are $3^7$ total possibilities for a string of 7 games. 3 of the 7 games need to be wins for A ${7 \choose 3}$, then 4 of the remaining games need to be draws ${4 \choose 2}$ and then the other two games are losses for B ${2 \choose 2}$. The total is then ${7 \choose 3}{4 \choose 2}{2 \choose 2}=210$ for part a.

(b) This should be the sum of the previous answer (210) and the total number of outcomes where A wins 4 games, ${7 \choose 4}$= 35. So, 245.

(c) For this one, I guess I just have to subtract from the answer for (b) because this includes the matches that go 7 games. I think there are two terms that need to be subtracted, ${6 \choose 4}{3 \choose 3}$, which denotes all outcomes where A wins four of the first 6 games and B wins the other 3. The other term ${6 \choose 3}{3 \choose 2}$, which is the result where A wins 3 of the first 6 and draws 2 of the remaining 3 games. So, the final answer would be $245-{6 \choose 4}{3 \choose 3}-{6 \choose 3}{3 \choose 2}=170$

Is it okay to subsume ${4 \choose 4}$ and ${5 \choose 4}$ in ${6 \choose 4}$? It seems like if I added these terms, I would be counting different outcomes two or three times.