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At Mike’s Hamburgers, each hamburger can be ordered with any of the fo

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At Mike’s Hamburgers, each hamburger can be ordered with any of the fo  [#permalink]

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New post 26 Nov 2019, 01:24
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A
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  65% (hard)

Question Stats:

50% (02:21) correct 50% (01:13) wrong based on 28 sessions

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Re: At Mike’s Hamburgers, each hamburger can be ordered with any of the fo  [#permalink]

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New post 26 Nov 2019, 01:51
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Bunuel wrote:
At Mike’s Hamburgers, each hamburger can be ordered with any of the following toppings: lettuce, tomato, onion, pickle, mayonnaise, mustard, ketchup, cheese. How many different combinations of toppings are possible?

A. 8
B. 56
C. 128
D. 256
E. 512


total possible ways ; 2^8 ; 256
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Re: At Mike’s Hamburgers, each hamburger can be ordered with any of the fo  [#permalink]

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New post 30 Nov 2019, 19:49
Can anybody explain this better ?

how did we get to decide the no of toppings we could choose.. starting 2 to ..7. or 8
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Re: At Mike’s Hamburgers, each hamburger can be ordered with any of the fo  [#permalink]

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New post 30 Nov 2019, 20:07
A burger can be ordered with any number of toppings - 0 to 8.
total ways possible = 8C0 + 8C1 + 8C2 + .... 8C8 = 256
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Re: At Mike’s Hamburgers, each hamburger can be ordered with any of the fo  [#permalink]

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New post 01 Dec 2019, 12:17
1. If the statement does not specify how many toppings we can choose, it means that we are free to choose from 0 to 8 toppings.

2. This means counting all possible combinations: N = 8; k =0,1,2,3...8 = 8C0 + 8C1 + 8C2... and so on

3. Instead of counting each combination from 0 to 8, one could note that we are simply counting the number of subsets that can be made out of 8 elements (toppings)

#of subsets of any set of elements A = \(2^n\) = \(2^8\) = 256



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Re: At Mike’s Hamburgers, each hamburger can be ordered with any of the fo   [#permalink] 01 Dec 2019, 12:17
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