There were two women participating in a chess | vedclass

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(C) Let the number of men be $n$. The total number of participants is $n + 2$.Each participant plays $2$ games with every other participant.The number

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) Let the number of men be $n$. The total number of participants is $n + 2$.Each participant plays $2$ games with every other participant.The number of games played between men is $2 	imes {^nC_2} = 2 	imes \frac{n(n-1)}{2} = n(n-1) = n^2 - n$.The number of games played between men and women is $2 	imes (n 	imes 2) = 4n$.According to the problem,the number of games between men exceeds the number of games between men and women by $66$:$n^2 - n - 4n = 66$$n^2 - 5n - 66 = 0$$(n - 11)(n + 6) = 0$Since $n$ must be positive,$n = 11$.The total number of participants is $n + 2 = 11 + 2 = 13$.

There were two women participating in a chess tournament. Every participant played two games with every other participant. The number of games that the men played between themselves exceeded the number of games that the men played with the women by $66$. The total number of participants is

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