(A) Let the number of men be $m$ and the number of women be $2$.Each participant plays $2$ games with every other participant.The number of games play

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(A) Let the number of men be $m$ and the number of women be $2$.Each participant plays $2$ games with every other participant.The number of games played between men is $2 \times \binom{m}{2} = 2 \times \frac{m(m-1)}{2} = m(m-1) = m^2 - m$.The number of games played between men and women is $2 \times (m \times 2) = 4m$.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,$m = 12$.

Class 11 Mathematics Permutations and Combinations Geometrical problems

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There are $m$ men and $2$ women participating in a chess tournament. Each participant plays $2$ 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

JEE MAIN 2019 All JEE MAIN

A
$12$
B
$11$
C
$9$
D
$7$

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There are $m$ men and $2$ women participating in a chess tournament. Each participant plays $2$ 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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