# 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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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) $\quad(1111)^{2}$