There are m men and two women participating in a chess tournament.

Question

There are $m$ men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of $m$ is:

Accepted Answer

Games among men: $2\binom{m}{2}=m(m-1)$.

Games between men and women: $2\times m\times2=4m$.

$$m(m-1) - 4m = 84 \implies m^2-5m-84=0 \implies (m-12)(m+7)=0 \implies m=12$$

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