A true-false quiz contains five questions. In how many ways can a student answer the question if the student answers two of the questions with “false” and the other three with “true”?

using polya strategy . answer with the step by step process step1 , understanding the problem. step 2. devise a plan, step3 carry out the plan number 4. review the solution . using simple english

### Step 1: Understanding the Problem

We are given a true-false quiz with 5 questions. The student answers exactly 2 questions with "false" and the other 3 questions with "true." We need to find how many different ways the student can answer the quiz. 

This is essentially a **combinatorics problem** because we are choosing 2 questions out of 5 to answer "false," and the remaining will automatically be "true."

### Step 2: Devise a Plan

Since we are dealing with combinations, the order in which the student chooses the answers does not matter. To solve this, we will use **combinations**, which allows us to count how many ways we can choose a specific number of items (questions in this case) from a set.

The number of ways to select 2 questions out of 5 to be "false" is given by the combination formula:

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

Where:
- \( n \) is the total number of questions (in this case, 5),
- \( r \) is the number of questions to be answered "false" (in this case, 2).

### Step 3: Carry Out the Plan

Using the combination formula to find how many ways to choose 2 questions to be "false":

\[
\binom{5}{2} = \frac{5!}{2!(5 - 2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]

So, there are **10 ways** for the student to select 2 questions to be "false," and the remaining 3 will automatically be "true."

### Step 4: Review the Solution

We have successfully calculated the number of ways to answer the quiz. The total number of ways the student can answer 2 questions with "false" and 3 with "true" is **10**. This makes sense, as we are simply choosing 2 out of 5 questions, and there are no other factors to consider (such as ordering).

The problem asked for the number of ways to answer the quiz under the given conditions, and we have found that there are 10 different ways.

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Do you want details on any step or have any questions? 

Here are 5 related questions to expand on this concept:
1. How would the result change if the quiz had 6 questions instead of 5?
2. Can we solve this problem using the binomial theorem?
3. How does this problem relate to probability?
4. What if the quiz allowed students to answer each question with "true," "false," or "skip"?
5. What are other examples of problems involving combinations?

**Tip:** When solving combinatorics problems, always identify whether order matters. If not, combinations (rather than permutations) are usually the way to go!

A true-false quiz contains five questions. In how many ways can a student answer the question if the student answers two of the questions with “false” and the other three with “true”?

Explore how to solve a combinatorics problem involving a 5-question true-false quiz where a student answers 2 questions with 'false' and the remaining 3 with 'true.' This problem demonstrates the use of the combination formula and is suitable for students in Grades 9-12.

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