Problem 19 - Precalculus with Limits

Question

In how many ways can a six-question true-false exam be answered? (Assume that no questions are omitted.)

Step-by-Step Solution

Answer

There are \( 2^6 = 64 \) ways to answer the six-question true-false exam.

Step 1: Identify the number of options for each event

For a true-false question, there are 2 possibilities: it can be answered either as true or as false.

Step 2: Count the total number of questions

In this case, there are a total of 6 questions to be answered.

Step 3: Apply the counting principle

By the counting principle, the number of ways to answer the exam is \( 2^6 \)

Key Concepts

Counting Principle

The Counting Principle is a fundamental concept in combinatorics that helps us determine the total number of outcomes for a sequence of events. Imagine you have a series of selections or tasks, and each one can be completed in a certain number of ways.

For instance, if you are answering a true-false exam with 6 questions, each question can be answered in 2 ways: either true or false. The Counting Principle tells us that to find the total number of ways to answer the exam, you multiply the number of options for each question. This is because each choice is independent of the others.

The formula given by the Counting Principle for our example is:

• Number of ways = number of options for question 1 × number of options for question 2 × ... × number of options for question n.

For 6 true-false questions, this becomes:

• 2 options per question times 6 questions = \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 = 64 \).

Thus, there are 64 different ways to answer the exam.

Permutations

Permutations are all about arranging or ordering a set of items. The concept helps us understand how many different ways we can organize a certain number of items. In permutations, the order of the items matters.

Let's imagine a scenario not directly in the exam problem but similar: arranging 3 different books on a shelf. How many ways can you do that? Use permutations here. Each position on the shelf can hold any of the books. So if you have 3 books, book A, book B, and book C:

• There are 3 choices for the first book.
• After placing the first book, 2 choices remain for the second.
• Finally, only 1 choice remains for the last spot.

This gives us a total of \( 3 \times 2 \times 1 = 6 \) ways to arrange the books. Represented as the permutation of 3 distinct books from a set of 3, written as \( 3! \), where \( n! \) or "n factorial" is the product of all positive integers up to n.

Permutations are particularly useful when you need to account for order, unlike combinations.

Combinations

Combinations are used when the order of the items does not matter. Unlike permutations, combinations consider only the selection and not the arrangement.

For instance, when you’re choosing 2 books out of 3, it doesn’t matter which order they are in. So, choosing book A and B is the same as choosing book B and A. When calculating combinations, you do not multiply by the number of arrangements.

• The formula for combinations is: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

where \( n \) is the total number of items you're choosing from, and \( r \) is the number of items to be selected.

Using the books example, if we want to choose 2 books out of 3, apply:

• \( \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \).

Thus, there are 3 ways to choose the books, and these would be (A, B), (A, C), and (B, C).

Combinations simplify problems where order doesn't affect the outcome.