Answer: E. 1200
Step-by-step explanation:
Given : A certain restaurant offers 8 different salads, 5 different main courses, 6 different desserts.
i.e. No. of choices for different salads= 8 -----(1)
No. of choices for different main courses = 5 ----(2)
No. of choices for different desserts = 6
When we choose two different desserts then we use permutation(repeatition not allowed) :
[tex]^6P_2=\dfrac{6!}{(6-2)!}=\dfrac{6\times5\times4!}{4!}=6\times5=30[/tex]----(3)
[∵ No. of ways to choose r things out of n =[tex]^nP_r=\dfrac{n!}{(n-r)!}[/tex] ]
If customers choose one salad, one main course and two different desserts for their meal , then By Fundamental principle of counting (Multiply (1) , (2) and (3)), the number of different meals are possible :-
[tex]8\times5\times30\\\\=1200[/tex]
Hence, the correct answer is E. 1200 .
