George is going on vacation and wishes to take along 2 books to read. If he has 5 different books to choose from, how many different combinations of 2 books can he bring ?

Options

A. 2
B. 5
C. 10
D. 15

Explanation

To find out how many different combinations of 2 books George can bring from 5 different books, we can use the concept of combinations in combinatorial mathematics.

### Step-by-Step Solution:

1. **Identify the total number of books**: George has 5 different books to choose from. Let's denote these books as A, B, C, D, and E.

2. **Determine the number of books to choose**: George wants to take 2 books on his vacation.

3. **Use the combination formula**: The number of ways to choose \( r \) items from \( n \) items without regard to the order of selection is given by the combination formula:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
where \( n \) is the total number of items, \( r \) is the number of items to choose, and \( ! \) denotes factorial.

4. **Plug in the values**: Here, \( n = 5 \) (the total number of books) and \( r = 2 \) (the number of books to choose).
\[ C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \cdot 3!} \]

5. **Calculate the factorials**:
- \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
- \( 2! = 2 \times 1 = 2 \)
- \( 3! = 3 \times 2 \times 1 = 6 \)

6. **Substitute the factorials back into the formula**:
\[ C(5, 2) = \frac{120}{2 \times 6} = \frac{120}{12} = 10 \]

7. **Conclusion**: Therefore, the number of different combinations of 2 books that George can bring is **10**.