Let there be x girls and y boys.
Number of games between two girls = $\binom{x}{2}$
Number of games between two boys = $\binom{y}{2}$
$\frac{x(x-1)}{2}$ = 45
$x^{2} - x$ – 90 = 0
(x– 10)(x + 9) = 0
x = 10
Similarly,
$\frac{y(y-1)}{2}$ = 190
$x^{2} - x$ – 380 = 0
(y + 19)(y – 20) = 0
y = 20
Total number of games = $\binom{x+y}{2}$ = $\binom{30}{2}$ = 435
Number of games in which one player is a boy and the other is a girl = 435 – 45 – 190 = 200
Hence, option (1)200 is the correct choice.