A certain restaurant offers 6 kinds of cheese and 2 kinds of

Question

A certain restaurant offers 6 kinds of cheese and 2 kinds of fruit for its dessert platter. If each dessert platter contains an equal number of kinds of cheese and kinds of fruit, how many different dessert platters could the restaurant offer?

A) 8
B) 12
C) 15
D) 21
E) 27

A. 8
B. 12
C. 15
D. 21
E. 27

Bunuel · Accepted Answer

As dessert platter should contain equal number of kinds of cheese and fruits, desert can contain:

A. 2 kinds of cheese and 2 kinds of fruits --> 6C2*2C2=15*1=15

B. 1 kind of cheese and 1 kind of fruit --> 6C1*2C1=6*2=12

A+B=15+12=27

Answer: E.

andrewcs · Answer

1 of each: 1 fruit * 1 cheese = 2 * 6 = 12 possibilities
2 of each: Well, you only have 2 fruits, so only one choice here. With 6 cheese, and 2 places, where order does not matter, you get 6!(6-2)!2! = 6!/4!2 = 6*5/2 = 15

So 15 + 12 = 27 total possible platters...

siddhans · Answer

How do we get 6c2 * 2c2?? Can someone break it down ? Do we read * as "AND" 2 cheese can be chosen from 6 cheese in 6c2 ways  AND   2fruit can be choosen from 2 in 2c2 ways hence whenver we have AND we multiply???

Bunuel · Answer

Yes, your understanding is correct.

It's called  Principle of Multiplication : if one event can occur in  m  ways and a second can occur independently of the first in  n  ways, then the two events can occur in  m*n  ways.

EgmatQuantExpert · Answer

This question is a good example where both OR and AND events are used. Let's see how such events are treated when finding the number of ways of an event happening.

Given Info
We are given that a restaurant in its dessert platter offers 6 kinds of cheese and 2 kinds of fruits. We are asked to find the number of ways a dessert platter can be offered if the platter contains equal kinds of fruits and cheese.

Approach
We observe here that there are lesser kinds of fruits than the cheese. Hence the number of ways the platter can be offered is limited by the number of fruits.

Since a platter would be incomplete without any of the fruit or cheese, for the event to occur( i.e. platter to be served) a platter should have fruit AND cheese.

Also, there can be a case where a platter may have 1 fruit AND 1 cheese OR 2 fruits AND 2 cheese. Hence, the event(i.e. platter to be served) can occur in either of the ways.

Number of ways the platter can be served = Number of ways of picking up 1 fruit out of 2 fruits AND number of ways of picking up 1 cheese out of 6 cheese OR Number of ways of picking up 2 fruits out of 2 fruits AND number of ways of picking up 2 cheese out of 6 cheese

Working Out
Number of ways in which 1 fruit can be picked out of 2 fruits = \(2_{c_{1}}\)

Number of ways in which 1 cheese can be picked out of 6 cheese = \(6_{c_{1}}\)

Hence number of ways in which a platter having 1 fruit and 1 cheese can be served = \(2_{c_{1}}\) \(*\) \(6_{c_{1}}\) \(= 2 * 6 = 12\)

Similarly, Number of ways in which 2 fruits can be picked out of 2 fruits = \(2_{c_{2}}\)

Number of ways in which 2 cheese can be picked out of 6 cheese = \(6_{c_{2}}\)

Hence number of ways in which a platter having 2 fruits and 2 cheese can be served = \(2_{c_{2}}\) \(*\) \(6_{c_{2}}\) \(= 1 * 15 = 15\)

Therefore total number of ways of serving a dessert platter \(= 12 + 15 = 27\)

Hope this helps

Regards
Harsh

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Regards Harsh

OptimusPrepJanielle · Answer

A certain restaurant offers 6 kinds of cheese and 2 kinds of fruit for its dessert platter. If each dessert platter contains an equal number of kinds of cheese and kinds of fruit, how many different dessert platters could the restaurant offer?

A) 8
B) 12
C) 15
D) 21
E) 27

A) 8
B) 12
C) 15
D) 21
 E) 27

Since the plates have to contain an equal number of cheese and fruit types, the only possibilities are 1 kind of cheese and 1 kind of fruit or 2 types of cheese and 2 types of fruit.

Case 1:
Cheese: n=6, r = 1 
 (6x5x4x3x2x1) 
1(5x4x3x2x1)

Fruit: n=2, r=1
 (2x1) 
1(1)

Combining cheese and fruit 6 x 2 =  12

Case 2:
Cheese n = 6, r = 2
 (6x5x4x3x2x1) 
2x1 (4x3x2x1)
=15

Fruit: There are only two fruits, so there is only one combination.

Combining cheese and fruits 15 x 1 =  15 
 
15 1fruit-1cheese combinations + 12 2fruit-2cheese combinations = 27 total combinations

ruis · Answer

Why would the following be wrong?? C(6,2)*C(2,2) + C(1,1)^6? I understand the explanation given above but I dont understand why this way of reasoning would not work. When should I square the combination? 
Another example is: what is the difference in meaning when we say 2*C (6,2) and C(6,2)^2. I usually mix up these....
I am studing the Math Book but always end up writing the solution with some kind of typo on my reasoning

Bunuel · Answer

6C2 * 2C2 means choosing 2 kinds of cheese  out of 6  and 2 kinds of fruits  out of 2 . However, what does C(1,1)^6 suppose to mean there?

kingbucky · Answer

Why have not considered the case where dessert platter is plain (0 kind of cheese and 0 kind of fruits) ?
Is it evident from the question stem ?

What would we mark as the answer if 28 were there as a answer choice, along with 27 ?

Bunuel · Answer

What kind of a dessert platter is that on the menu with nothing on it? Don’t overthink it.

bumpbot · Answer

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A certain restaurant offers 6 kinds of cheese and 2 kinds of

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