A company produces 8 different types of candies, and sells the candies in gift packs. How many different gift packs containing exactly 3 different candy types can the company put on the market?

Options

A. `._(8)C_(2)`
B. `._(8)C_(3)`
C. `._(8)C_(2)`
D. `._(8)P_(3)`

Answer

To solve the problem of how many different gift packs containing exactly 3 different candy types can be made from 8 different types of candies, we can use the concept of combinations.

### Step-by-Step Solution:

1. **Identify the total number of candy types (n)**:
The company produces 8 different types of candies. Therefore, \( n = 8 \).

2. **Identify the number of candy types to choose (r)**:
We need to select 3 different types of candies for each gift pack. Therefore, \( r = 3 \).

3. **Use the combination formula**:
The number of combinations of \( n \) items taken \( r \) at a time is given by the formula:
\[ nCr = \frac{n!}{r!(n - r)!} \]
Here, \( n! \) denotes the factorial of \( n \).

4. **Substitute the values into the formula**:
We need to calculate \( 8C3 \):
\[ 8C3 = \frac{8!}{3!(8 - 3)!} = \frac{8!}{3! \cdot 5!} \]

5. **Calculate the factorials**:
- \( 8! = 8 \times 7 \times 6 \times 5! \) (we can cancel \( 5! \) in the numerator and denominator)
- \( 3! = 3 \times 2 \times 1 = 6 \)

6. **Simplify the expression**:
\[ 8C3 = \frac{8 \times 7 \times 6}{3!} = \frac{8 \times 7 \times 6}{6} \]
The \( 6 \) in the numerator and denominator cancels out:
\[ 8C3 = 8 \times 7 = 56 \]

7. **Final answer**:
The total number of different gift packs containing exactly 3 different candy types is \( 56 \).

### Summary:
The company can produce **56 different gift packs** containing exactly 3 different types of candies.