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This is a homework question, but I've clearly made a mistake that I can't point out...

Nadal and Federer play in a grand slam tournament. They have to win three sets to win. (Best of five). How many possible games are there?

There are 2 options with only three sets: NNN and FFF.

I figured that for the options with 4 sets, e.g. NNFN or FNFF.
I did 4C3 - 2, because NNNF and FFFN are not possible. 4C3 - 2 = 2

Then there are also the games with 5 sets, like NNFFN or FNFNF.
I figured that this should be 5C3 - 4, because NNNFF, FNNNF, FFFNN, NFFFN are not possible. This gives me 6.

All of this added up is 11. The answer book said 20. I can't figure out where I messed up.

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  • $\begingroup$ Why is the number of games finite? A set has to be won by $2$ games. If the players trade victories they can play forever. Isner/Mahut went to $183$ games. $\endgroup$ – lulu Nov 1 '15 at 15:10
  • $\begingroup$ In the case 4, you forgot $FNNN$ and $NFFF$ $\endgroup$ – mrprottolo Nov 1 '15 at 15:10
  • $\begingroup$ @lulu I think the problem was poorly worded. It looks like we are counting the number of possible arrangements of sets. $\endgroup$ – N. F. Taussig Nov 1 '15 at 15:11
  • $\begingroup$ @N.F.Taussig Oh, of course. thanks. $\endgroup$ – lulu Nov 1 '15 at 15:14
  • $\begingroup$ FNNN and NFFF are possible, since the game stops after 3 sets won by the same person. NNNF continues after three games won by Nadal, and is not a possible arrangement. $\endgroup$ – Eowyn12 Nov 1 '15 at 15:19
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To count the games of length $4$, let us count the games of length $4$ where F wins, and multiply by $2$.

To win in $4$, $F$ has to win the last set, and two of the first three sets. Which two? They can be chosen in $\binom{3}{2}$ ways. Double. We get $6$.

Foe length $5$, again count the ways F can win and double. F has to win the fifth set, and two of the first four. Which two? They can be chosen in $\binom{4}{2}$ ways. When we double we get $12$.

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I believe the intent of the question is to cover all possible scenarios, and the ans of $20$ is correct.

To start with, suppose Nadal wins: Instead of counting ways of winning in $3/4/5$ sets,

realize that he has to win $3$ sets placed somewhere in $5$, hence $\dbinom53$ = 10 ways.

Double it to cover the possibilities of Federer winning

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