It's \(V\) versus \(X\) and \(Y\)

Consider 2 numbers \(X\) and \(Y\) which are chosen at random from the set \(V=\{1,2,\ldots,3n\}\) without replacement . If the probablity that \(X^{3}+Y^{3}\) is divisible by 3 can be written of the form \[\dfrac{n^{a}}{n^{b}+2^{c}},\]

where \(a,b,c\in\mathbb{Z}\) and \(n\in\mathbb{N}\), find \((a+c)\times b\)

If you feel that the probability cannot be expressed in this form then type 22 as your answer.

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