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In a treasure chest are contained five different flawed diamonds and

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In a treasure chest are contained five different flawed diamonds and  [#permalink]

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New post 11 Sep 2019, 07:18
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A
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In a treasure chest are contained five different flawed diamonds and three different perfect diamonds. If a pirate selects four diamonds from the chest, in how many ways can four diamonds be selected, if exactly two of them must be perfect diamonds?

A - 3
B - 10
C - 20
D - 30
E - 60
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Re: In a treasure chest are contained five different flawed diamonds and  [#permalink]

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New post 11 Sep 2019, 07:34
Choosing exactly 2 out of 5 flawed diamonds = 5C2 = 10
Choosing exactly 2 out of 3 perfect diamonds = 3C2 = 3
Total ways = 10*3 = 30
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Re: In a treasure chest are contained five different flawed diamonds and  [#permalink]

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New post 04 Dec 2019, 21:14
Thanks MahmoudFawzy

I'm wondering why it's wrong to use 3C1 as the numerator. Wouldn't that account for the number of ways to choose 1 perfect diamond out of the 3 selections that the pirate makes? eg if there are 4 girls and 5 boys in a class, the probability that 1 girl is chosen out of a selection of 3 children is 3C1. LMK what I'm missing here.
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Re: In a treasure chest are contained five different flawed diamonds and  [#permalink]

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New post 05 Dec 2019, 01:00
This is a fairly simple question on the concept of combinations. The language of the question is also not too complicated since it uses the term ‘selection’ explicitly, which is a clue for you to understand that we need to use the \(n_c_r\) formula to solve this question.

The treasure chest contains a total of 8 diamonds of which 5 are distinct and flawed and the remaining 3 are distinct and perfect. We need to select a total of 4 diamonds of which exactly 2 of them must be perfect diamonds. This means that the other two should be flawed diamonds.

The number of ways of selecting ANY 2 perfect diamonds from 3 distinct perfect diamonds = \(3_C_2\) = 3 ways

The number of ways of selecting ANY 2 perfect diamonds from 5 distinct perfect diamonds = \(5_C_2\) = 10 ways.

Since we are supposed to perform both the tasks simultaneously, we say “ We have to select 2 perfect diamonds AND also select 2 flawed diamonds.”. The word AND is associated with the multiplication operator (this is the multiplication principle).

Therefore, total number of ways of selecting 4 diamonds in such a way that exactly 2 are perfect = Number of ways of selecting 2 perfect diamonds AND Number of ways of selecting 2 flawed ones = \(3_C_2\) x \(5_C_2\) = 3 x 10 = 30 ways.
The correct answer option is D.

Hope that helps!
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Re: In a treasure chest are contained five different flawed diamonds and   [#permalink] 05 Dec 2019, 01:00
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