In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

Options

A. 200
B. 216
C. 235
D. 256

Answer

To solve the problem, we need to determine the number of games played where one player is a boy and the other is a girl. We are given the number of games played by girls and boys separately.

### Step-by-Step Solution:

1. **Define Variables**:
Let \( n \) be the number of boys and \( m \) be the number of girls.

2. **Games Played by Girls**:
The number of games played between girls is given by the combination formula \( \binom{m}{2} \):
\[ \binom{m}{2} = 45 \]
This can be expressed as:
\[ \frac{m(m-1)}{2} = 45 \]
Multiplying both sides by 2:
\[ m(m-1) = 90 \]

3. **Solve for \( m \)**:
We need to find two consecutive integers whose product is 90. The pairs are:
- \( 9 \times 10 = 90 \)
Thus, \( m = 10 \) (the number of girls).

4. **Games Played by Boys**:
The number of games played between boys is given by:
\[ \binom{n}{2} = 190 \]
This can be expressed as:
\[ \frac{n(n-1)}{2} = 190 \]
Multiplying both sides by 2:
\[ n(n-1) = 380 \]

5. **Solve for \( n \)**:
We need to find two consecutive integers whose product is 380. The pairs are:
- \( 19 \times 20 = 380 \)
Thus, \( n = 20 \) (the number of boys).

6. **Calculate Games with One Boy and One Girl**:
The number of games where one player is a boy and the other is a girl can be calculated as:
\[ n \times m = 20 \times 10 = 200 \]

### Final Answer:
The number of games in which one player was a boy and the other was a girl is **200**.

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