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03 Mar 2022, 06:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
58% (01:57) correct
42%
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wrong
based on 493
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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)
Are You Up For the Challenge: 700 Level Questions
(A) 8!
(B) (3)(7!)
(C) (3)(8!)
(D) (8)(2^7)
(E) (3)(2^7)
Are You Up For the Challenge: 700 Level Questions
03 Mar 2022, 07:04
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Choosing meat: Each customer has a choice of 3
Choosing condiments: Each customer can have anywhere from zero to all condiments. Easiest way to look at this is that each condiment can either be a 'yes' or a 'no'. Which means that for every condiment there are two choices. As there are 7 condiments, \(2*2*2*2*2*2*2\) or \(2^7\)
Total different burgers possible: \(3*2^7\)
Answer E
Choosing condiments: Each customer can have anywhere from zero to all condiments. Easiest way to look at this is that each condiment can either be a 'yes' or a 'no'. Which means that for every condiment there are two choices. As there are 7 condiments, \(2*2*2*2*2*2*2\) or \(2^7\)
Total different burgers possible: \(3*2^7\)
Answer E
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03 Mar 2022, 16:53
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Out of the 7 condiments one may choose as many or as few, hence the total number of combinations of condiments available = 7C0 + 7C1 + 7C2 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7
From the math property we know that sum of nCr where r ranges from [0, n] = 2^n
Therefore an individual has 2^7 combination choices from the condiments. But we know that there are 3 varieties of meat as well. And each variety may be clubbed with any of these 2^7 combinations of condiments.
Hence, total different kinds of burgers possible = 3*(2^7)
Hence, E.
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From the math property we know that sum of nCr where r ranges from [0, n] = 2^n
Therefore an individual has 2^7 combination choices from the condiments. But we know that there are 3 varieties of meat as well. And each variety may be clubbed with any of these 2^7 combinations of condiments.
Hence, total different kinds of burgers possible = 3*(2^7)
Hence, E.
Posted from my mobile device
