BurgerTown offers many options for customizing a burger. There are 3

Question

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)

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Official Answer

E

(A) 8!
(B) (3)(7!)
(C) (3)(8!)
(D) (8)(2^7)
(E) (3)(2^7)

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Nidzo · Accepted Answer

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

G-Man · Answer

Hi  Bunuel

The question says 'A burger must include meat'
But it doesn't limit customers to only one kind of meat. Based on the question stem, a customer could select all 3 types of meat (however unhealthy it is :p)
In that case, won't there be 7 ways (3C1 + 3C2 + 3C3) to select meat?
Please clarify.

Regards

kungfury42 · Answer

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   = 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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Bhavita. · Answer

Hello All,

The question here says 'A burger must include meat', however it doesn't limit customers to only one kind of meat. Thus shouldn't it be the case where he can select any kind/number of meat (number >=1)
In that case, won't there be 7 ways (3C1 + 3C2 + 3C3) to select meat?
Please clarify.

Regards

kingbucky · Answer

The question could be made clearer; I thought that the customer can have multiple different types of meat.

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Adarsh_24 · Answer

Yeah this question should have mentioned. Customer can have only 1 type of meat in addition to must.
As it is answer should be 7*2^7.

Usernamevisible · Answer

## GMAT counting pattern:

Step 1: Choose the meat
There are 3 types of meat.
Choices = 3
-----------

Step 2: Choose condiments
There are 7 condiments.
For each condiment, you have 2 choices:
• Include it
• Do not include it
So total condiment combinations:
2^7
This already includes:
• no condiments
• exactly 1 condiment
• all 7 condiments
• any combination in between
----------------------------

Step 3: Multiply independent choices
3 × 2^7
= 3 × 128
= 384
Therefore:
(3)(2^7)
Answer: (E)
-----------

GMAT shortcut
Whenever items are optional, think:
Each item contributes a factor of 2 (yes/no).
So 7 optional condiments → 2^7.
Then multiply by any required choices (the 3 meats).

BurgerTown offers many options for customizing a burger. There are 3

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BurgerTown offers many options for customizing a burger. There are 3

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