There are two women participating with some men in a chess tournament. Each participant played two games…
There are two women participating with some men in a chess tournament. Each participant played two games with the other. The number of games that the men played between themselves is 66 more than that of the men played with the women. Then the total number of participants in the tournament is
$17$
$13$
$11$
$19$
Solution
Let there be n men participants
Then the number of games that the men play between themselves is $2{ }^n C_2$
And the number of games that the men played with women is $2 \times 2 n$
$\therefore 2{ }^n C_2-2 \times 2 n=66 \Rightarrow \frac{2 n!}{(n-2)!2!}-4 n=66$
$\Rightarrow n^2-5 n-66=0 \Rightarrow n=11$ or -6 (rejected)
Hence, total number of participants $=11+2=13$