8 fiction novels by new authors are on display at a Book Festival. Patrons are told that they can choose any 3 books. How many different selections can a patron make?

The problem involves choosing 3 books from a set of 8 books. Since the order in which the books are chosen does not matter, this is a combination problem.

The formula for combinations is:

\[
C(n, r) = \frac{n!}{r!(n - r)!}
\]

Where:
- \( n \) is the total number of items (in this case, 8 books),
- \( r \) is the number of items to be chosen (in this case, 3 books).

Substituting the values into the formula:

\[
C(8, 3) = \frac{8!}{3!(8 - 3)!} = \frac{8!}{3!5!}
\]

Simplify the factorials:

\[
C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56
\]

So, there are **56 different selections** a patron can make.

Would you like more details or have any questions? Here are five related questions to explore:

1. How would the solution change if patrons could select 4 books instead of 3?
2. How do combinations differ from permutations in this context?
3. What if there were 10 books on display, and patrons still choose 3?
4. How does the formula for combinations change if the number of books to choose equals the total number of books?
5. In how many ways can the books be selected if patrons must choose books in a specific order?

**Tip:** Combinations are used when the order of selection doesn't matter, while permutations are for when the order does.

Explore how to calculate combinations using the formula C(n, r) for a problem involving 8 books and choosing 3. This guide breaks down the formula and shows step-by-step how to simplify the factorials to find that there are 56 different selections. Suitable for Grades 9-12 students learning combinations and basic probability.

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