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jrymbei
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4 flavours * 4 choices = 4C1*4C1 = 4*4=16

Note: The question says in that among the options of cheese, mushroom or both, they have the choice but not the obligation of picking the option. So we have a 4 options - cheese, mushroom, both or *none*.
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In this case, if it becomes obligatory to add either cheese, mushroom, or both.
Then the "none" varieties will be 0.

Total varieties would then be 12.
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Hi All,

From the answer choices, we know that the number of total possibilities is limited, so we can 'map them' all out:

4 types of pizza and 4 options per pizza:

Pepperoni - as is, extra cheese, mushrooms, BOTH cheese/mushrooms = 4 options
Chicken - as is, extra cheese, mushrooms, BOTH cheese/mushrooms = 4 options
Hawaiian - as is, extra cheese, mushrooms, BOTH cheese/mushrooms = 4 options
Vegetarian - as is, extra cheese, mushrooms, BOTH cheese/mushrooms = 4 options

Total = 4+4+4+4 = 16 varieties

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jrymbei
A university cafeteria offers 4 flavors of pizza - pepperoni, chicken, Hawaiian and vegetarian. If a customer has an option (but not the obligation) to add extra cheese, mushrooms or both to any kind of pizza, how many different pizza varieties are available ?

(A) 4
(B) 8
(C) 12
(D) 16
(E) 32

We are given 4 flavors for pizza and there are 4 options for each flavor: no toppings, cheese-only, mushrooms-only, and both toppings. Thus, we have 4 x 4 = 16 different pizzas possible.

Answer: D
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Hi
i am a bit confused why wouldnt the answer be 4*4*4*4 when to multiply and when to add
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Gmat4321prep
Hi
i am a bit confused why wouldnt the answer be 4*4*4*4 when to multiply and when to add

As you would have noticed scrolling through the answer all of them calculated separately the different variety of pizzas for a particular flavor(pepperoni, chicken, Hawaiian, and vegetarian).


Flavor total variety
1. pepperoni 4
2. chicken 4
3. Hawaiian 4
4. vegetarian 4

therefore, 4+4+4+4=16

solve more and more question you will certainly able to figure out where to use multiplication or addition.
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I went with something like 4C1 *2C1 *2C1 *2C2 = 16
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