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 :-
[tex]^nC_r=\dfrac{n!}{(n-r)!r!}[/tex]
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
= [tex]^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84[/tex]
No. of ways to answer 4 open-ended problems
= [tex]^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15[/tex]
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
= [tex]84\times15=1260[/tex]
Hence, the correct answer is option A) 1260
