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 6 of the multiple-choice questions and 4 of the open-ended problems, in how many ways can the questions and problems be chosen?

A) 1260
B) 1296
C) 261,273,600
D) 21,772,800

Answer 1

Answer: A) 1260

Explanation:

We know that the number of combinations of n things taking r at a time is given by :-

`^nC_r=(n!)/((n-r)!r!)`

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

=
`^9C_6=(9!)/(6!(9-6)!)=(9*8*7*6!)/(6!3!)=84`

No. of ways to answer 4 open-ended problems

=
`^6C_4=(6!)/(4!(6-4)!)=(6*5*4!)/(4!2!)=15`

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

=
`84*15=1260`

Hence, the correct answer is option A) 1260