Question

A catering service offers 5 appetizers,4 main courses, and 8 desserts. A customer is to select 4 appetizers,2 main courses,and 3 desserts for a banquet. In how many ways can this be done

Answer 1

Answer:

468 ways

Step-by-step explanation:

Given: A catering service offers 5 appetizers, 4 main courses, and 8 desserts

To find: number of ways a customer is to select 4 appetizers, 2 main courses,and 3 desserts.

Solution:

A permutation is an arrangement of elements such that order of elements matters and repetition is not allowed.

Number of appetizers = 5

Number of main courses = 4

Number of desserts = 8

Number of ways of choosing k terms from n terms = $nPk=\frac{n!}{(n-k)!}$

Number of ways a customer is to select 4 appetizers, 2 main courses,and 3 desserts =  $5P4+4P2+8P3$

$=\frac{5!}{(5-4)!}+\frac{4!}{(4-2)!}+\frac{81}{(8-3)!}\\=5!+\frac{4!}{2!}+\frac{8!}{5!}\\=5!+(4\times 3)+(8\times 7\times 6)\\=120+12+336\\=468$

So, this can be done in 468 ways.