Question

There were two women participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves proved to exceed by 66 the number of games that the men played with the women. How many participants were there in the tournament and how many games were played?

Answer 1

Let number of man participating in tournament $=n$

number of women participating in tournament $=2$

Each player will play $2$ games with every players.

$\therefore$ Two woman will play two sames with each man.

Total games played by woman $=2\times \left(2n\right)$

$=4n$

Total number of games played by man among Themselves $=\left(Totalnumberofpairsofpossibleman\right)\times 2$ each pair will play

so $n(n-1)-4n=66$

$n^2-5n-66=0$

$n^2-11n+6n-66=0$

$(n-11)(n+6)=0$

$n=11$

Total number of participants $=man+woman$

$=11+2=13$

Total games played $=\left(pairsofplayers\right)\times 2$

${}^{13}{C}_2\times 2=\frac{13\times 12}{2}\times 2$

$=156$