There are m men and two women participating i | vedclass

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(A) Let the number of men be $m$ and the number of women be $2$. Total participants = $m + 2$.Each participant plays $2$ games with every other partic

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) Let the number of men be $m$ and the number of women be $2$. Total participants = $m + 2$.Each participant plays $2$ games with every other participant.The number of games played between men themselves is given by $2 	imes ^mC_2 = 2 	imes \frac{m(m-1)}{2} = m(m-1) = m^2 - m$.The number of games played between men and women is given by $2 	imes (m 	imes 2) = 4m$.According to the problem,the difference between these is $84$:$m^2 - m - 4m = 84$$m^2 - 5m - 84 = 0$$(m - 12)(m + 7) = 0$Since $m$ must be positive,$m = 12$.

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

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