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you choose 1 appetizer

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by sanju09 » Tue Oct 18, 2011 5:31 am
For dinner at a restaurant, there are x choices of appetizers, y + 1 main courses, and z choices of dessert. How many total possible choices are there if you choose 1 appetizer, 1 main course, and 1 dessert for your meal?
(A) x + y + z + 1
(B) x y z + x z
(C) x y + z + 1
(D) x y z + 1
(E) x y z + ½
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by leonswati » Tue Oct 18, 2011 5:48 am
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
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by neelgandham » Tue Oct 18, 2011 6:23 am
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
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by zooki » Tue Oct 18, 2011 6:34 am
Your possible choices are:
xC1*(y+1)C1*zC1
=x*(y+1)*z
=xz(y+1)
=xyz+xz
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