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Question

The value of (21C110C1)+(21C210C2)+(21C310C3)+(21C410C4)+...+(21C1010C10) is

A
221211
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B
221210
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C
22029
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D
220210
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Solution

The correct option is D 220210
(21C110C1)+(21C210C2)+(21C310C3)+...+(21C1010C10)
=(21C1+21C2++21C10)(10C1+10C2++10C10)=(21C1+21C2++21C10)(2101)=12(21C1+21C2+21C211)(2101)=12(2212)(2101)=2201210+1=220210

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