Discussion
There are two women participants in a badminton tournament. The number of games the men played between themselves exceeds by 12 the number of games they played with women. If each player played one game with each other, then the number of men in the tournament was
Answer: D
There were 8 men.
Let there be \(m\) men. Games among the men number \(\binom{m}{2}\), and games between men and the two women number \(2m\).
\[ \begin{aligned} \frac{m(m - 1)}{2} - 2m &= 12 \\ m^{2} - m - 4m &= 24 \\ m^{2} - 5m - 24 &= 0 \\ (m - 8)(m + 3) &= 0 \end{aligned} \]
So \(m = 8\). Check: 28 games among the men, 16 against women, and \(28 - 16 = 12\). With 6 men the counts would be 15 and 12, a difference of only 3.
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