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Re: Esteban’s restaurant offers a lunch special. A customer can [#permalink]
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sandy wrote:
Explanation

This problem is about combinations, because the order of the dishes does not matter.

Since you’re choosing 4 dishes, start by drawing 4 blanks. On top, write the number of choices: 12 choices for the first dish, then 11, 10, and 9. On the bottom, start with the size of the smaller group and count down: 4, 3, 2, and 1.

Cancel the numbers on the bottom, and the numbers on top will multiply to 495.

Hence option C is correct!


Can you please explain in more detail?
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Re: Esteban’s restaurant offers a lunch special. A customer can [#permalink]
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This a classic combinatorics problem, in which you count the 4 dishes from 12 (the order does not matter) and apply the classic formula.

\(\frac{12*11*10*9*8*7*6*5*4*3*2*1}{(8*7*6*5*4*3*2*1)*(4*3*2*1)} =\)

Simplify

\(3*11*5*3 = 495\)

C is the answer
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Re: Esteban’s restaurant offers a lunch special. A customer can [#permalink]
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Esteban’s restaurant offers a lunch special. A customer can [#permalink]
Carcass wrote:
This a classic combinatorics problem, in which you count the 4 dishes from 12 (the order does not matter) and apply the classic formula.

\(\frac{12*11*10*9*8*7*6*5*4*3*2*1}{(8*7*6*5*4*3*2*1)*(4*3*2*1)} =\)

Simplify

\(3*11*5*3 = 495\)

C is the answer



Sir how do we sort out which problem is based in permutation which in combination?
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Re: Esteban’s restaurant offers a lunch special. A customer can [#permalink]
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A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:

I pick 2 persons among 20...any kind


A permutation is an ordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as:


two persons among 20 but first the tallest and then the shortest....


https://gre.myprepclub.com/forum/gre-quant ... 18819.html
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Estebans restaurant offers a lunch special. A customer can [#permalink]
Esteban’s restaurant offers a lunch special. A customer can order a platter consisting of four different small dishes from a selection of twelve choices. How many different platters can a customer create?

From a choice of 12 different dishes, he can choose four.

Since the order of dishes does not matter, it is a combination problem.

So, the number of platters that customer can create is \(12C4 \)

\(12C4 = 12!/(12-4)! * 4! = 12!/8! * 4! = 12 * 11 * 10 * 9/4! = 495 \)

The Answer is C.
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