Question

There were two participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves proved to exceed by $66$ the number of games that the men played with the women. The number of participants is?

• A. $6$
• B. $11$
• C. $13$
• D. None of these

Answer 1

The correct option is C

$13$

Explanation for the correct option:

Find the total number of participants

If $n$ is the number of men participating in the game, then number of games played by men among themselves is $2\times {{}^{n}C}_2$

Number of games played by men with women is $2\times 2n$

Thus, as per the given condition,

$$\begin{gathered} 2\times {{}^{n}C}_2-2\times 2n=66 \\ \Rightarrow 2\frac{n!}{\left(n-2\right)!2!}-4n=66 \\ \Rightarrow n\left(n-1\right)-4n=66 \\ \Rightarrow n^2-5n-66=0 \\ \Rightarrow \left(n-11\right)\left(n+6\right)=0 \end{gathered}$$

$n=-6$ is not possible so $n=11$

Therefore, total number of participants is equal to $11$(men)$+$ $2$(women) $=13$

Hence, the correct option is $\left(\mathrm{C}\right)$