Step 1: Understanding the Question:
The problem describes a round-robin chess tournament where everyone plays everyone else exactly once.
We are given the number of matches played exclusively between two girls (45 games) and exclusively between two boys (190 games).
We need to find the total number of cross-gender matches (one boy playing against one girl).
Step 2: Key Formula or Approach:
If there are $n$ people, the number of matches between them is given by combinations: $^nC_2 = \frac{n(n-1)}{2}$.
Let the total number of girls be $g$ and the total number of boys be $b$.
We will set up two equations: $\frac{g(g-1)}{2} = 45$ and $\frac{b(b-1)}{2} = 190$ to find the exact values of $g$ and $b$.
The number of games played between a boy and a girl is simply the product of the number of boys and the number of girls ($b \times g$).
Step 3: Detailed Explanation:
Let $g$ denote the total number of girls participating in the competition.
The number of games where both players are girls is $^gC_2$.
We are given that $^gC_2 = 45$.
Using the formula: $\frac{g(g-1)}{2} = 45$.
Multiplying both sides by 2 gives $g(g-1) = 90$.
We look for two consecutive integers that multiply to 90. Since $10 \times 9 = 90$, there are exactly $g = 10$ girls.
Let $b$ denote the total number of boys participating in the competition.
The number of games where both players are boys is $^bC_2$.
We are given that $^bC_2 = 190$.
Using the formula: $\frac{b(b-1)}{2} = 190$.
Multiplying both sides by 2 gives $b(b-1) = 380$.
We look for two consecutive integers that multiply to 380. Since $20 \times 19 = 380$, there are exactly $b = 20$ boys.
A boy vs girl game is formed by selecting 1 boy out of $b$ and 1 girl out of $g$.
The number of such mixed games is simply $b \times g$.
Total boy vs girl games = $20 \times 10 = 200$.
Step 4: Final Answer:
The number of games in which one player was a boy and the other was a girl is 200.