Problem 93 - Algebra and Trigonometry

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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 \( C(10, 8) = 10! / [8!(10-8)!] = 45 \).

Step 2: 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 \( C(5, 3) = 5! / [3!(5-3)!] = 10 \).

Step 3: 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 = \( C(10, 8) * C(5, 3) = 45 * 10 = 450 \).