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Question

The value of (300)(3010)(301)(3011)+(302)(3012)++(3020)(3020)


Solution

The correct option is D
(1x)30=30C0x030C1x1+30C2x2+(1)3030C30x30 .............(i)
(1+x)30=30C0x30+30C1x29+30C2x28+30C10x0++30C30x0 .............. (ii) 
Multiplying (i) and (ii) and equating the coefficient of x20 on both sides, we get required sum = coefficient of x20 in (1x2)30=30C10

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