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

English

Gujarati

Hindi

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

Share

Similar Questions

A car will hold $2$ in the front seat and $1$ in the rear seat. If among $6$ persons $2$ can drive, then the number of ways in which the car can be filled is

medium

If all the letters of the word $'GANGARAM'$ be arranged, then number of words in which exactly two vowels are together but no two $'G'$ occur together is-

normal

If $^{43}{C_{r – 6}} = {\,^{43}}{C_{3r + 1}},$ then the value of $r$ is

easy

The least value of natural number $n$ satisfying $C(n,\,5) + C(n,\,6)\,\, > C(n + 1,\,5)$ is

easy

A group of students comprises of $5$ boys and $n$ girls. If the number of ways, in which a team of $3$ students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is $1750$, then $n$ is equal to

hard

(JEE MAIN-2019)