Maths Olympiad Prep

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Combinatorics Difficulty 5.2 AIME, harder Find the answer

In a table tennis single round-robin tournament with 15 athletes, each athlete plays against every other athlete once. If Athlete 1 wins x1x_{1} matches, Athlete 2 wins x2x_{2} matches, ‥ and Athlete 15 wins x15x_{15} matches, find the value of x1+x2+⋯+x15x_{1}+x_{2}+\cdots+x_{15}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Seven, according to the problem, each athlete competes in 14 matches. Therefore, Athlete 1 wins x1x_{1} matches and loses (14−x1)\left(14-x_{1}\right) matches;
Athlete 2 wins x2x_{2} matches and loses (14−x2)\left(14-x_{2}\right) matches;
\qquad
Athlete 15 wins x15x_{15} matches and loses (14−x15)\left(14-x_{15}\right) matches.
Thus, we have
x1+x2+⋯+x16=(14−x1)+(14−x2)+⋯+(14−x15). \begin{array}{l} x_{1}+x_{2}+\cdots+x_{16} \\ =\left(14-x_{1}\right)+\left(14-x_{2}\right)+\cdots+\left(14-x_{15}\right) . \end{array}

Solving this, we get
x1+x2+⋯+x15=105.  x_{1}+x_{2}+\cdots+x_{15}=105 \text {. }

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.