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olga nikolaevna [1]
3 years ago
11

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
Mathematics
1 answer:
katrin [286]3 years ago
5 0

Answer: A) 1260

Step-by-step explanation:

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

^nC_r=\dfrac{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=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84

No. of ways to answer 4 open-ended problems

= ^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{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\times15=1260

Hence, the correct answer is option A) 1260

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