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Use Polya's four-step method in problem solving to solve. If a test has four true/false questions, in how many ways can there be three answers that are false and one answer that is true?

Short Answer

Expert verified
Hence, there are 4 possible ways in which there can be exactly three false answers and one true.

Step by step solution

01

Identifying the values of n and k

From the problem, we know that there are 4 true/false questions in total. This makes our 'n', the total number of items, 4, and since we want to find the combinations with 3 false answers, our 'k', the number of items to choose, is 3.
02

The Combinations Formula

Now we use the combinations formula: \( C(n,k) = \frac{n!}{k!(n-k)!} \). Plugging the values \( n=4 \) and \( k=3 \) into the formula, we get \( \frac{4!}{3!(4-3)!} \).
03

Computing the Factorial Values and the Final Result

We know that \(4! = 4*3*2*1 = 24\), \(3! = 3*2*1 = 6\) and \((4-3)! = 1\). So now we substitute these values into our formula which will look like this \( C(n,k) = \frac{24}{6*1} = 4 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polya's Problem-Solving Method
George Polya, a mathematician, introduced a simple, four-step approach to tackle problems methodically. This approach is not limited to math but is useful across various disciplines; it fosters critical thinking and a systematic way to address challenges. Let's apply Polya’s method to the exercise provided.

  • Understand the problem: The first step is to comprehend what is being asked. In our exercise, we are required to determine in how many ways three out of four answers can be false on a true/false test.
  • Devise a plan: Next, we formulate a strategy to solve the problem, which, in this case, involves using combinatorics, specifically the combinations formula.
  • Carry out the plan: Execute the chosen strategy. We will calculate the combinations where 'n' is the total number of questions and 'k' is the number of false answers desired.
  • Look back: Finally, review the solution to see if it makes sense and if there's a possibility for simplification or an alternative method.

By systematically following these steps, students can efficiently tackle even the most daunting problems.
Factorial Notation
Factorial notation is a mathematical concept used to describe the product of an integer and all the integers below it down to one. For instance, the factorial of 4 (denoted as '4!') is calculated as
\( 4! = 4 \times 3 \times 2 \times 1 = 24 \).

Factorials are pivotal in combinatorics, especially in calculating permutations and combinations. Zero factorial,
\( 0! \), is defined to be 1, which is an important convention for dealing with scenarios where no items are chosen from a set. Understanding how to compute factorial values prepares one for tackling a variety of combinatorial problems.
Combinatorics
Combinatorics is a branch of mathematics concerned with counting, arrangement, and combination of elements within sets, according to specific rules. It's pivotal in probability, computer science, and problem-solving. In the context of our problem, combinatorics helps in calculating how many ways one can select 'k' items from a set of 'n' items without considering the order, which is exactly what the combinations formula calculates.

The combinations formula, which is represented as \( C(n, k) = \frac{n!}{k!(n-k)!} \), shows how many ways 'k' items can be picked from a set of 'n', ignoring the sequence. This formula helps to determine all possible groups that can be formed from a larger set and is a foundational concept in combinatorial mathematics.
Probability
Probability is the measure of how likely an event is to occur out of the total number of possible outcomes. It plays a crucial role in statistics, risk assessment, and everyday decision making. Mathematically, it's expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

While probability wasn't directly applied in solving our example problem, it's often interlinked with combinatorics. Understanding how many possible combinations exist allows one to calculate the probability of a particular event occurring. For instance, if we want to find the probability of getting exactly three false answers on a true/false test with four questions, we would divide the number of desired combinations (in our case, 4) by the total number of possible combinations, which would involve calculating the power of 2 to the 4th power, since there are two possibilities (true or false) for each of the four questions.

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Most popular questions from this chapter

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