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A round-robin tournament is being held with n tennis players; this means 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). How many games are played in total?

I would think that are $\dfrac{n(n-1)}{2}\ 2^{n} $ possible outcomes, with $\dfrac{n(n-1)}{2}\ $ games played in total, is this correct?

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  • $\begingroup$ You have the elements of the right reasoning. Can you see that it is $2^g$, where $g=\frac{n(n-1)}{2}$? $\endgroup$ – André Nicolas Nov 24 '14 at 18:10
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You’ve misused the multiplication rule. There are $\dfrac{n(n-1)}2$ games, and each has $2$ possible outcomes, so there are

$$\large2^{\frac{n(n-1)}2}$$

possible outcomes.

For example, with $n=3$ there are $3$ games, and each has $2$ possible outcomes, so there are $2\cdot2\cdot2=2^3$ possible results for the tournament.

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  • $\begingroup$ Worth noting: N(N-1)/2 = (N choose 2) (which is also the same as the sum of the first N-1 numbers) $\endgroup$ – Learning stats by example Mar 20 '20 at 17:01

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