Question

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen?

Answer 1

Answer:

720 ways

Step-by-step explanation:

Generally, combination is expressed as;

                                  $^{n} C_{r} = \frac{n!}{r!(n-r)!}$

The question consists of 9 multiple-choice questions and examinee must answer 7 of the multiple-choice questions.

                                   ⇒ ⁹C₇ $=\frac{9!}{7!(9-7)!}$

                                        $=\frac{9!}{7!(2)!}$

                                        = 36

The question consists of 6 open-ended problems and examinee must answer 3 of the open-ended problems.

                                    ⇒ ⁶C₃ $=\frac{6!}{3!(6-3)!}$

                                         $=\frac{6!}{3!(3)!}$

                                         = 20

Combining the two combinations to determine the number of ways the questions and problems be chosen if an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problem.

                                        ⁹C₇ × ⁶C₃

                                       = 36 × 20

                                       = 720 ways