Attached is a question from GMAT Prep Test #2.
What is the most efficient way of achieving the result?
Answer: E
Thanks,
K
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GMAT Prep #2_DS Combinatorics #34
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Bill@VeritasPrep
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Since we need an equal number of cheeses and fruits, we have the following possibilities:
1) 1 cheese and 1 fruit
We are choosing 1 cheese from 6 options and 1 fruit from 2 options
= 6 cheeses * 2 fruits
=12 platters
2) 2 cheeses and 2 fruits
We are choosing 2 cheeses from 6, and there is only 1 way to choose both fruits.
=6!/(2!4!) * 1
=(6*5)/2 * 1
=15
15 + 12 = 27 different platters.
1) 1 cheese and 1 fruit
We are choosing 1 cheese from 6 options and 1 fruit from 2 options
= 6 cheeses * 2 fruits
=12 platters
2) 2 cheeses and 2 fruits
We are choosing 2 cheeses from 6, and there is only 1 way to choose both fruits.
=6!/(2!4!) * 1
=(6*5)/2 * 1
=15
15 + 12 = 27 different platters.
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It is given that each dessert platter contains an equal number of cheese and fruit. So, the possible combinations are:kwah wrote:Attached is a question from GMAT Prep Test #2.
What is the most efficient way of achieving the result?
Answer: E
Thanks,
K
(1) 2 kinds of cheese and 2 kinds of fruits
Number of ways of choosing 2 kinds of cheese and 2 kinds of fruits = 6C2 * 2C2 = 6!/(2! * 4!) * 2!/2! = (6 * 5)/2 = 15
(2) 1 kind of cheese and 1 kind of fruit
Number of ways of choosing 1 kind of cheese and 1 kind of fruits = 6C1 * 2C1 = 6 * 2 = 12
Therefore, number of different dessert platters that the restaurant can offer = 15 + 12 = 27
The correct answer is E.
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