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)

Are You Up For the Challenge: 700 Level Questions

Show Answer

Official Answer

E

(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

G-Man · Accepted 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

Nidzo · 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

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.

Posted from my mobile device

sayan640 · Answer

MartyMurray  Would you like to discuss this question ?

Posted from my mobile device

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.

Posted from my mobile device

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.

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards