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For dinner at a restaurant, there are x choices of appetizers, y + 1

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

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New post 09 Jun 2016, 03:46
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Question Stats:

82% (01:07) correct 18% (01:30) wrong based on 133 sessions

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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

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

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New post 09 Jun 2016, 04:03
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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


x appetizers,
y + 1 main courses,
z dessert

Required: 1 appetizer, 1 main course, and 1 dessert
Number of ways possible = xC1*(y+1)C1*zC1 {NC1 = N! / (N-1)!*1! = N}

Hence, number of ways = x(y+1)z = xyz + xz

Correct 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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New post 09 Jun 2016, 04:22
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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New post 09 Jun 2016, 09:07
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]

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New post 09 Mar 2019, 04:50
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Re: For dinner at a restaurant, there are x choices of appetizers, y + 1   [#permalink] 09 Mar 2019, 04:50
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