Two women and some men participated in a ches | vedclass

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

Vedclass · Accepted Answer

$(11, 13]$

Vedclass · Answer

(C) Let the number of men be $n$.Total participants $= n + 2$.Each participant plays $2$ games with every other participant.The number of games played between $n$ men is $2 	imes \binom{n}{2} = 2 	imes \frac{n(n-1)}{2} = n(n-1)$.The number of games played between $n$ men and $2$ women is $n 	imes 2 	imes 2 = 4n$ (since each man plays $2$ games with each of the $2$ women).Given that the number of games between men exceeds the number of games between men and women by $66$:$n(n-1) - 4n = 66$$n^2 - n - 4n = 66$$n^2 - 5n - 66 = 0$$(n - 11)(n + 6) = 0$Since $n > 0$,we have $n = 11$.The value $n = 11$ lies in the interval $(11, 13]$.

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