Ways to choose one appetizer = XC1 = X

Ways to choose one main course = (Y+1)C1 = Y+1

Ways to choose one dessert = ZC1 = Z

Ways to choose one appetizer, one main course and one dessert = X*(Y+1)*Z =XYZ + XZ

Answer B

An alternate method, if you are stuck and if you don't know what to do.

Let x =1, y+1 =1, z = 1, then the # of ways of serving = 1. substituting the values

(A) x + y + z + 1 = 4 (Not the answer)
(B) x y z + x z = 1 (May be)
(C) x y + z + 1 = 2 (Not the answer)
(D) x y z + 1 = 1 (May be)
(E) x y z + ½ = 1/2 (Not the answer)

We now have two options to choose from,

Let x =1, y+1 =1, z = 2, then the # of ways of serving = 2. substituting the values

(B) x y z + x z = 2 is the answer
(D) x y z + 1 = 1 Nopes

Answer B

Your possible choices are:
xC1*(y+1)C1*zC1
=x*(y+1)*z
=xz(y+1)
=xyz+xz