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BurgerTown offers many options for customizing a burger. The

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GMAT Club Legend
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Joined: 07 Jun 2014
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GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1600 [2] , given: 373

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BurgerTown offers many options for customizing a burger. The [#permalink] New post 30 Jul 2018, 10:44
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Question Stats:

42% (00:44) correct 57% (00:43) wrong based on 7 sessions
BurgerTown offers many options for customizing a burger. There are 3 types of meats and 7 condiments: lettuce, tomatoes, pickles, onions, ketchup, mustard, and special sauce. A burger must include meat, but may include as many or as few condiments as the customer wants. How many different burgers are possible?

(A) \(8!\)
(B) \((3)(7!)\)
(C) \((3)(8!)\)
(D) \((8)(2^7)\)
(E) \((3)(2^7)\)
[Reveal] Spoiler: OA

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Joined: 21 Jun 2018
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Re: BurgerTown offers many options for customizing a burger. The [#permalink] New post 05 Aug 2018, 21:37
sandy wrote:
BurgerTown offers many options for customizing a burger. There are 3 types of meats and 7 condiments: lettuce, tomatoes, pickles, onions, ketchup, mustard, and special sauce. A burger must include meat, but may include as many or as few condiments as the customer wants. How many different burgers are possible?

(A) 8!
(B) (3)(7!)
(C) (3)(8!)
(D) (8)(27)
(E) (3)(27)


Answer is (3)(28). Please confirm and provide explanation.
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jun 2014
Posts: 4704
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 90

Kudos [?]: 1600 [0], given: 373

CAT Tests
Re: BurgerTown offers many options for customizing a burger. The [#permalink] New post 07 Aug 2018, 06:09
Expert's post
Explanation

This problem tests the fundamental counting principle, which states that the total number of choices is equal to the product of the independent choices. The key to this problem is realizing how many choices there are for each option. For the meat, there are 3 choices.

For each of the condiments there are exactly 2 choices: yes or no. The only real choice regarding each condiment is whether to include it at all. As there are 7 condiments, the total number of choices is \((3)(2)(2)(2)(2)(2)(2)(2) = (3)(2^7)\).

Note: the condiment options cannot be counted as 8! (0 through 7 = 8 options) because, in this case, the order in which the options are chosen does not matter; a burger with lettuce and pickles is the same as a burger with pickles and lettuce.
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Re: BurgerTown offers many options for customizing a burger. The   [#permalink] 07 Aug 2018, 06:09
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