There are m men and two women participating in a chess tournament. Each participant plays two games

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6.Permutation and Combination

hard

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

A

$12$

B

$11$

C

$9$

D

$7$

(JEE MAIN-2019)

Solution

Let $m-$ men, $2-$ women

$^m{C_2} \times 2{ = ^m}{C_1}^2{C_1}.2 + 84$

${m^2} – 5m – 84 = 0$ $ \Rightarrow (m – 12)(m + 7) = 0$

$m = 12$

Standard 11

Mathematics

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