If in a chess tournament each contestant plays once against each of the other and in all 45 games are played, then the number of participants, is

Options

A. 9
B. 10
C. 15
D. none of these

Answer

To solve the problem, we need to determine the number of participants in a chess tournament where each contestant plays once against each other, and a total of 45 games are played.

### Step-by-Step Solution:

1. **Understanding the Problem**:
Each contestant plays against every other contestant exactly once. If there are \( n \) contestants, the number of games played can be represented by the combination formula \( \binom{n}{2} \), which gives the number of ways to choose 2 contestants from \( n \).

2. **Setting Up the Equation**:
We know that the total number of games played is 45. Therefore, we can set up the equation:
\[ \binom{n}{2} = 45 \]
This can be expanded to:
\[ \frac{n(n-1)}{2} = 45 \]

3. **Eliminating the Fraction**:
To eliminate the fraction, multiply both sides of the equation by 2:
\[ n(n-1) = 90 \]

4. **Rearranging the Equation**:
Rearranging gives us a quadratic equation:
\[ n^2 - n - 90 = 0 \]

5. **Factoring the Quadratic**:
We need to factor the quadratic equation. We look for two numbers that multiply to -90 and add to -1. The numbers -10 and 9 work:
\[ (n - 10)(n + 9) = 0 \]

6. **Finding the Values of \( n \)**:
Setting each factor to zero gives us:
\[ n - 10 = 0 \quad \Rightarrow \quad n = 10 \]
\[ n + 9 = 0 \quad \Rightarrow \quad n = -9 \quad (\text{not valid since } n \text{ must be positive}) \]

7. **Conclusion**:
The only valid solution is \( n = 10 \). Therefore, the number of participants in the chess tournament is 10.

### Final Answer:
The number of participants is \( \boxed{10} \).