The total number of games played in chess tournament 
There are two women 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.Find the total number of games played 
  in tournament.

$a.)\ 132\\
b.)\ 112\\
c.)\ 156\\
\color{green}{d.)\ 210} $
Let the number of men be $x$.
$2\dbinom{x}{2}=4x+66 \\
\implies x=11 $
Total games played = $2\dbinom{11+2}{2}=156$.
But the answer given in book is $d.)\ 210 .$
I look for a short and simple way.
I have studied maths up to $12$th grade.
 A: Your translation of the problem is not correct.  Letting $x$ be the number of men and $y$ the number of women, then the men played $2{x \choose 2} = x(x-1)$ games among themselves.  The men played $2xy$ games with the women, so you need $x(x-1)=2xy+66$  If you multiply by $\frac 1x$ you get $x-1=2y+\frac {66}x$, so $x$ must divide $66$.  Try the factors and see what you get.
A: Well, you may look it as a graph $G=(V,E)$ fully connected with some $V$ which is the number of players.
Let $E$ be the relations between players.
Let $E'$ be the relations between the men.
Let also $C$ be the games between men and women.
We simply note that we form $ E = E' + C + 1 $, since the edges are the ones connecting the men, the edges between the men and women and the edge between both women. We know that $ C = 2(V-2) $ as well as $ E = \frac{V(V-1)}{2} $. Solving returns $ V = 13 $. The total relations we have is $ E = \frac{(13) \  (12)}{2} $.
Then the number of games is $ \text{games} = (13)(12) = 156 $.

The image shows the graph. The women are the nodes $A$ and $D$. The group of edges $C$ is the now shown. The missing edge between the women is omitted because it isn't considered when comparing $C$ and $E'$.

Tell me if you think I made an error somewhere.

Checking back (because I doubted it too), let's try with $210 \text{ games}$.
$$
210 \text{ games} \implies V = 15
$$
So working with that:
$$
C = 2(15-2) = 26 \\
E = E' + C + 1 = (26 + 33) + 26 + 1 = 86 \implies V \neq 15
$$
