Suppose we have a tennis tournament with $n$ players so that every player will play against every other player exactly once. How many possible outcomes are there for the tournament? (the outcome lists out who won and who lost for each game)


first, first player can play with $n-1$ other players, the second player plays with $n-2$ remaining players and so on. So, in total we have

$$ 1 + 2 + 3 + ... + n-2 + n-1 = \frac{n (n-1) }{2} $$

games played. So, for each game we only have two possibilities. either one wins or other loses so that is

$$ \underbrace{2 \times 2 \times 2 \times \cdots \times 2}_{n(n-1)/2 \; \text{times}} = 2^{ \frac{ n(n-1) }{2}}$$

Is this a correct solution?

  • $\begingroup$ Looks right to me. Good job. $\endgroup$
    – saulspatz
    Apr 6, 2018 at 2:30

1 Answer 1


Yes, it is correct.

You may interpret the question as finding all combinations of selecting 2 from n people, which is $\binom{n}{2}=\frac{n!}{(n-2)!2!}=\frac{n(n-1)}{2}$

Again, you may take each game as flipping a coin. Therefore, the number of possible outcomes for flipping a coin $\frac{n(n-1)}{2}$ times is $2^{\frac{n(n-1)}{2}}$.


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