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.
_________________

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...

E.
In case we serve 1 of each 6C1*2C1=6*2=12
In case we serve 2 of each 6C2*2C2=15*1=15
=> 15+12=27

.

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???

Answer:
Option E = 27.

Explanation:

For 1 item each of Fruit & Cheese : 2 x 6 = 12

For 2 item each: Both fruit go in the dish. And we have to select 2 cheese out of 6 i.e. 6-C-2 = 6!/(4! x 2!) = 15

Total Menu Items = 15 +12 = 27

1) One of Each, Therefore 2*6

2) Two of Each,But we have two fruits anyway.So that`ll be one. Rest two Cheese variants can be chosen from the 6 types in 6 C 2 ways.
6! / (4! * 2!)

Total of 1 & 2 gives 27.

6C1*2C1 + 6C2*2C2 = 27

the platter could contain either 2 items or 4 items.
2-items: \(6*2 = 12\)

4-items: \(\frac{6*5}{2!}*\frac{2*1}{2!} = 15\)

total = 15+12 = 27

E

Two possibilities, either 2 types of each or 1 type of each.

For 2 types: 6C2 * 2C2 = 15 * 1 = 15
For 1 type: 6C1 * 2C1 = 6 * 2 = 12

Total is 15+12 = 27.

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.
_________________

6+6+6c2=27.
E is the answer .
The above expression denoted choosing one cheese type and one fruit type (6+6) ways ,for each type of fruit there are 6 types of cheese that can be selected .
And finally where there are two types of cheese and fruits respectively .
_________________

I got e as well

For 2 types: 6C2 * 2C2 = 15 * 1 = 15
For 1 type: 6C1 * 2C1 = 6 * 2 = 12

exact approach

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
_________________

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

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________