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Chapter 11: Problem 93

A mathematics exam consists of 10 multiple-choice questions and 5 open-ended problems in which all work must be shown. If an examinee must answer 8 of the multiple-choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen?

Short Answer

Expert verified
The student can choose the questions to answer in 450 different ways.

Step by step solution

01

Calculate the combinations for multiple-choice questions

Calculate the number of ways to choose 8 questions out of 10. This can be done using the combination formula: \( C(n, k) = n! / [k!(n-k)!] \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose. Here, \( n=10 \) and \( k=8 \). So, the number of ways to choose 8 questions out of 10 is \( C(10, 8) = 10! / [8!(10-8)!] = 45 \).
02

Calculate the combinations for open-ended problems

Calculate the number of ways to choose 3 problems out of 5. Similar to the previous calculation, using the combination formula, where \( n=5 \) and \( k=3 \), the number of ways to choose 3 problems out of 5 is \( C(5, 3) = 5! / [3!(5-3)!] = 10 \).
03

Multiplication Principle of Counting

Now, to find the total possible ways to pick 8 multiple-choice questions and 3 open-ended problems, multiply the number of ways to do each task separately. This follows the multiplication principle of counting, which states that if there are \( m \) ways to do one task and \( n \) ways to do another, then there are \( m*n \) ways to do both. Thus, total ways = \( C(10, 8) * C(5, 3) = 45 * 10 = 450 \).

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Most popular questions from this chapter

Use this information to solve Exercises \(47-48 .\) The mathematics department of a college has 8 male professors, 11 female professors, 14 male teaching assistants, and 7 female teaching assistants. If a person is selected at random from the group, find the probability that the selected person is a professor or a female.

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