Question

A certain restaurant offers 8 different salads, 5 different main courses, 6 different desserts. If customers choose one salad, one main course and two different desserts for their meal, how many different meals are possible?
A. 120
B. 240
C. 480
D. 600
E. 1200

Answer 1

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) :

$^6P_2=\dfrac{6!}{(6-2)!}=\dfrac{6\times5\times4!}{4!}=6\times5=30$----(3)

[∵ No. of ways to choose r things out of n =$^nP_r=\dfrac{n!}{(n-r)!}$ ]

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 :-

$8\times5\times30\\\\=1200$

Hence, the correct answer is E. 1200 .