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I did this slightly differently,

6c2*1/3*1/3*4c2*1/3*1/3*2c2*1/3*1/3 = 10/81

Bunuel , is this correct?
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There are 6 spaces and we have A, B, and C with their probability as P(A) = P(B) = P(C) = \(\frac{1}{3}\)

=> First selection and twice occuring: \(^6{C_2} * \frac{1}{3} * \frac{1}{3} = \frac{30 }{ 18}\)

=> Second selection and twice occuring: \(^4{C_2} * \frac{1}{3} * \frac{1}{3} = \frac{12 }{ 18}\)

=> Third selection and twice occuring: \(^2{C_2} * \frac{1}{3} * \frac{1}{3} = \frac{2 }{ 18}\)

=> \(\frac{30 }{18} * \frac{12 }{ 18} * \frac{2 }{ 18} = \frac{10 }{ 81}\)

Answer C
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Hello from the GMAT Club BumpBot!

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