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A pizza can be assembled from given choices, with customer’s preferenc

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A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 29 May 2018, 01:48
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Difficulty:

  65% (hard)

Question Stats:

54% (02:02) correct 46% (01:59) wrong based on 165 sessions

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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

To solve question 3: P&C Practice Question 3

Related Articles:

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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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 29 May 2018, 05:59
EgmatQuantExpert wrote:
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


EgmatQuantExpert - Is there a mistake in the question stem?
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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 29 May 2018, 06:02
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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 29 May 2018, 06:08
1
EgmatQuantExpert wrote:
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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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 29 May 2018, 06:26
EgmatQuantExpert wrote:
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 wrote:
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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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 31 May 2018, 01:52

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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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 01 Jun 2018, 07:02
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we can also try one more method.

Since we have an option of choosing (not choosing ) any cheese or topping.We can take this as extra option.

eg. now no. of toppings =5 (including the no choice option)
now of types of cheese=4(including the no choice option)

so, the total ways=3x5x4= 60
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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 11 Jan 2019, 22:10
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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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 12 May 2019, 03:39
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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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 25 May 2019, 06:40
EgmatQuantExpert wrote:

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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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 25 Aug 2019, 18:44
EgmatQuantExpert wrote:
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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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 26 Aug 2019, 10:16
EgmatQuantExpert wrote:
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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Re: A pizza can be assembled from given choices, with customer’s preferenc  [#permalink]

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New post 26 Aug 2019, 11:30
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 wrote:
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

Related Articles:

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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Re: A pizza can be assembled from given choices, with customer’s preferenc   [#permalink] 26 Aug 2019, 11:30
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