A basketball term has 5 centers ,9 guards and 13 forwards .Of these , 1 center , 2 guards , and 2 forwards start a game .How many possible starting terms can a coach put on the floor ?

Options

A. 56160
B. 14040
C. 585
D. 197

Explanation

To solve the problem of how many possible starting teams a coach can put on the floor, we need to calculate the combinations of players selected from each category: centers, guards, and forwards.

### Step-by-Step Solution:

1. **Identify the number of players to be selected**:
- We need to select 1 center from 5 available centers.
- We need to select 2 guards from 9 available guards.
- We need to select 2 forwards from 13 available forwards.

2. **Calculate the combinations for each category**:
- For centers: The number of ways to choose 1 center from 5 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).
\[ C(5, 1) = \frac{5!}{1!(5-1)!} = \frac{5!}{1! \cdot 4!} = \frac{5 \cdot 4!}{1 \cdot 4!} = 5 \]

- For guards: The number of ways to choose 2 guards from 9 is:
\[ C(9, 2) = \frac{9!}{2!(9-2)!} = \frac{9!}{2! \cdot 7!} = \frac{9 \cdot 8}{2 \cdot 1} = 36 \]

- For forwards: The number of ways to choose 2 forwards from 13 is:
\[ C(13, 2) = \frac{13!}{2!(13-2)!} = \frac{13!}{2! \cdot 11!} = \frac{13 \cdot 12}{2 \cdot 1} = 78 \]

3. **Combine the results**:
- Since we need to select players from different categories simultaneously, we multiply the number of combinations:
\[ \text{Total combinations} = C(5, 1) \times C(9, 2) \times C(13, 2) = 5 \times 36 \times 78 \]

4. **Calculate the total**:
- First, calculate \( 5 \times 36 \):
\[ 5 \times 36 = 180 \]
- Next, multiply this result by 78:
\[ 180 \times 78 = 14,040 \]

5. **Final Answer**:
- The total number of possible starting teams the coach can put on the floor is **14,040**.