Math Problem Statement

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?

Solution

This problem can be solved using combinations since the order in which the books are selected doesn't matter. The formula for combinations is given by:

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

Where:

  • nn is the total number of items (in this case, books),
  • rr is the number of items to be chosen.

Here:

  • n=8n = 8 (total books),
  • r=3r = 3 (books to be selected).

Substituting these values into the combination formula:

C(8,3)=8!3!(83)!=8!3!5!C(8, 3) = \frac{8!}{3!(8 - 3)!} = \frac{8!}{3!5!}

Simplifying the factorials:

C(8,3)=8×7×63×2×1=3366=56C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56

So, the number of different selections a patron can make is 56.

Would you like any further details or explanations?

Here are 5 related questions to think about:

  1. How would the solution change if the patron could only choose 2 books instead of 3?
  2. What if there were 10 books to choose from instead of 8?
  3. How many selections could a patron make if they were allowed to choose up to 4 books?
  4. What if the patron could select the same book more than once (allowing repetitions)?
  5. How would this change if the order of selection mattered (permutations)?

Tip: Remember that combinations are used when the order of selection doesn't matter, while permutations are used when order does matter.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Combinations

Formulas

C(n, r) = n! / (r!(n - r)!)

Theorems

Combinatorial Selection Theorem

Suitable Grade Level

Grades 9-12