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

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Bunuel
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At Mike’s Hamburgers, each hamburger can be ordered with any of the fo [#permalink] New post   26 Nov 2019, 01:24
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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
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Re: At Mike’s Hamburgers, each hamburger can be ordered with any of the fo [#permalink] 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
IMO D
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Re: At Mike’s Hamburgers, each hamburger can be ordered with any of the fo [#permalink] 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] 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] New post   01 Dec 2019, 12:17
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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



IMO D
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Re: At Mike’s Hamburgers, each hamburger can be ordered with any of the fo [#permalink] New post   27 Dec 2021, 23:57
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Since each burger can be ordered with any of the 8 toppings. He can choose 0 toppings or 1 topping 0r... .. or 8 toppings for the burger.

Total no of ways = 8C0 + 8C1 + 8C2 + 8C3 + 8C5 + 8C6+ 8C7 + 8C8

We know that nC0 + nC1+ nC2+ ... + nCn = \(2^n\)

Applying the above property here , the total no of ways = 2^8 = 256

Option D is the answer.

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Clifin J Francis,
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Re: At Mike’s Hamburgers, each hamburger can be ordered with any of the fo [#permalink]
27 Dec 2021, 23:57
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