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A boy have 32 cards out of which 10 cards, each of blue, green and red colours, have denominations as \( \left\{2^{1}, 2^{2}, \ldots, 2^{10}\right\} \) and one black and one white each has value unity. The number of ways in which the boy can get a sum of 2012 if he can choose any number of cards, is (A) \( (1001)^{2} \) (B) \( (1002)^{2} \) (C) \( (1007)^{2} \) (D) \( \quad(1111)^{2} \)

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