Problem 93 - College Algebra

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 total number of ways to choose the questions and problems can be calculated by multiplying the number of combinations from step 1 and step 2.

Step 1: Calculate combinations for multiple-choice questions

First, calculate the number of combinations for choosing 8 questions out of 10. Use the combination formula mentioned earlier substituting \( n = 10 \) and \( k = 8 \): \( C(10, 8) = \frac{10!}{8!(10-8)!} \)

Step 2: Calculate combinations for open-ended problems

Next, calculate the number of combinations for choosing 3 problems out of 5. Using the combination formula, substitute \( n = 5 \) and \( k = 3 \): \( C(5, 3) = \frac{5!}{3!(5-3)!} \)

Step 3: Calculate the total ways

The total ways will be the product of combinations from step 1 and step 2, because these are independent events (choosing multiple-choice questions does not affect choosing open-ended problems). So, multiply results from step 1 and step 2.