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EgmatQuantExpert - Is there a mistake in the question stem?

Hi,
Can you specify what mistake you are indicating?
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A pizza can be assembled from given choices, with customers' preferences. The crust is available in 3 varieties – cheese burst, classic, and wheat thin; toppings are available in 4 variants – bacon, black olives, pepperoni and sausage; and finally cheese available in 3 types – Cheddar, American, and Swiss. How many various types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese?

A. 3
B. 9
C. 12
D. 36
E. 60

The highlighted portion - I'm not so sure, but the meaning is not clear.
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A pizza can be assembled from given choices, with customers' preferences. The crust is available in 3 varieties – cheese burst, classic, and wheat thin; toppings are available in 4 variants – bacon, black olives, pepperoni and sausage; and finally cheese available in 3 types – Cheddar, American, and Swiss. How many various types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese?

pushpitkc
The highlighted portion - I'm not so sure, but the meaning is not clear.

This indicates while making a choice, a customer may choose one of the types of toppings and/or one of the types of cheese. However he may also choose not to include any toppings or cheese as his choice.
Hope this clears. :-)
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Solution



Given:
    • Number of varieties for
      o Pizza crust = 3
      o Toppings = 4
      o Cheese = 3
    • One have the choice of not choosing either of toppings or cheese

To find:
    • Number of ways pizza can be customized

Approach and Working:
There are 4 possible cases:
    • Choice of only crust, which can be done in \(^3C_1\) = 3 ways
    • Choice of crust and toppings, which can be done in \(^3C_1 * ^4C_1\) = 3 * 4 = 12 ways
    • Choice of crust and cheese, which can be done in \(^3C_1 * ^3C_1\) = 3 * 3 = 9 ways
    • Choice of crust, toppings, and cheese, which can be done in \(^3C_1 * ^4C_1 * ^3C_1\) = 3 * 4 * 3 =36 ways
Therefore, the total number of ways pizza can be customized = 3 + 12 + 9 + 36 = 60 ways

Hence, the correct answer is option E.

Answer: E
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Hi,

has the choice of not choosing either of toppings or cheese?

has the choice of not choosing toppings OR not choosing cheese .
Hence he needs to choose one of those .

Choice of crust and toppings = 3*4 = 12 ways
Choice of crust and cheese = 3*3 = 9 ways
Choice of crust, toppings, and cheese = 3*4*3 = 36 ways

Answer must be 57 ways
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Number of varieties for
Pizza crust = 3
Toppings = 4
Cheese = 3
One have the choice of not choosing either of toppings or cheese
SO 1 VERITY OF CRUST WITH 1 VERITY OF TOPPINGS WILL HAVE 4 CHOICES(3 CHEESE AND 1 ADDED BKZ SOMEONE CAN EVEN CHOOSE NO CHEESE).
SIMILARLY 1 VERITY OF CRUST WITH 4 VERITY OF TOPPINGS WILL HAVE 20 CHOICES(5*4).5 BKZ SOMEONE CAN EVEN CHOOSE NO TOPPINGS ATALL.
BUT SINCE CRUST IS MANDATORY,TOTAL COMBINATIONS POSSIBLE=3*5*4=60
OA-E
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Solution



Given:
    • Number of varieties for
      o Pizza crust = 3
      o Toppings = 4
      o Cheese = 3
    • One have the choice of not choosing either of toppings or cheese

To find:
    • Number of ways pizza can be customized

Approach and Working:
There are 4 possible cases:
    • Choice of only crust, which can be done in \(^3C_1\) = 3 ways
    • Choice of crust and toppings, which can be done in \(^3C_1 * ^4C_1\) = 3 * 4 = 12 ways
    • Choice of crust and cheese, which can be done in \(^3C_1 * ^3C_1\) = 3 * 3 = 9 ways
    • Choice of crust, toppings, and cheese, which can be done in \(^3C_1 * ^4C_1 * ^3C_1\) = 3 * 4 * 3 =36 ways
Therefore, the total number of ways pizza can be customized = 3 + 12 + 9 + 36 = 60 ways

Hence, the correct answer is option E.

Answer: E

Hi,
I dont understand the 36 ways concept. why we are taking all the variants here?
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EgmatQuantExpert
A pizza can be assembled from given choices, with customers' preferences. The crust is available in 3 varieties – cheese burst, classic, and wheat thin; toppings are available in 4 variants – bacon, black olives, pepperoni and sausage; and finally cheese available in 3 types – Cheddar, American, and Swiss. How many various types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese?

A. 3
B. 9
C. 12
D. 36
E. 60


The customer has 3 choices of crust, 5 choices of toppings (either one of the 4 toppings or none) and 4 choices of cheese (either one of the 3 types or none). Thus, a total of 3 x 5 x 4 = 60 different pizzas can be ordered.

Answer: E
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EgmatQuantExpert
A pizza can be assembled from given choices, with customers' preferences. The crust is available in 3 varieties – cheese burst, classic, and wheat thin; toppings are available in 4 variants – bacon, black olives, pepperoni and sausage; and finally cheese available in 3 types – Cheddar, American, and Swiss. How many various types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese?

A. 3
B. 9
C. 12
D. 36
E. 60


Given:
1. A pizza can be assembled from given choices, with customers' preferences.
2. The crust is available in 3 varieties – cheese burst, classic, and wheat thin;
3. toppings are available in 4 variants – bacon, black olives, pepperoni and sausage;
4. and finally cheese available in 3 types – Cheddar, American, and Swiss.

Asked: How many various types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese?

Choices for crust = 3
Choices for toppings = 4 + 1 = 5
Choices for cheese = 3 + 1 = 4

Number of types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese = 3*5*4 = 60

IMO E
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The language of this question is very confusing.
Where is it written that their can not be two toppings on a pizza or two types of cheese ? or three toppings ?




EgmatQuantExpert
A pizza can be assembled from given choices, with customers' preferences. The crust is available in 3 varieties – cheese burst, classic, and wheat thin; toppings are available in 4 variants – bacon, black olives, pepperoni and sausage; and finally cheese available in 3 types – Cheddar, American, and Swiss. How many various types of pizzas can be made if a customer has the choice of not choosing either of toppings or cheese?

A. 3
B. 9
C. 12
D. 36
E. 60

To solve question 3: P&C Practice Question 3

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Article-1: Learn when to “Add” and “Multiply” in Permutation & Combination questions
Article-2: Fool-proof method to Differentiate between Permutation & Combination Questions
Article-3: 3 deadly mistakes you must avoid in Permutation & Combination

All Articles: Must Read Articles and Practice Questions to score Q51 !!!!
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Another way:

work out the initial combinations:
3 ways to choose crust, 4 ways to choose toppings and 3 ways to choose cheese
total combinations = \(3*4*3\) = 36 ways

since more options are provided, there must be >36 ways, and the only answer is E
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