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Re: For dinner at a restaurant, there are x choices of appetizers, y + 1 [#permalink]
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Kudos
We have to choose 1 appetizer, 1 main course, and 1 dessert for the meal out of x choices of appetizers, y + 1 main courses, and z choices of dessert.

This can be done in xC1 . (y+1)C1 . zC1 ways = x(y+1)z = xyz + xz (Option B).
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Re: For dinner at a restaurant, there are x choices of appetizers, y + 1 [#permalink]
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Bunuel wrote:
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. xyz + xz
C. xy + z + 1
D. xyz + 1
E. xyz + 1/2


Possible ways to choose 1 appetizer out of x= x
Possible ways to choose 1 main course out of y+1= y+1
Possible ways to choose 1 dessert out of z= z

So possible ways to choose 1 appetizer, 1 main course, and 1 dessert:-

x (y+1) (z) = xyz+xz

B is the answer
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Re: For dinner at a restaurant, there are x choices of appetizers, y + 1 [#permalink]
4
Kudos
Let X,y,z each be 2
Y;3 and X,z are both 2
Total choices of ordering one of each 2*3*2;12
Put value in option
Only b matches with 12 as (2*2*2+2*2)




Bunuel wrote:
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. xyz + xz
C. xy + z + 1
D. xyz + 1
E. xyz + 1/2


Posted from my mobile device
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Re: For dinner at a restaurant, there are x choices of appetizers, y + 1 [#permalink]
5
Kudos
Choose 1 appetizer, 1 main course, and 1 dessert for the meal out of x choices of appetizers, y + 1 main courses, and z choices of dessert.

=> xC1 . (y+1)C1 . zC1 ways = x(y+1)z = xyz + xz

Hence B
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Re: For dinner at a restaurant, there are x choices of appetizers, y + 1 [#permalink]
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Re: For dinner at a restaurant, there are x choices of appetizers, y + 1 [#permalink]
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