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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