The value of (21C1−10C1)+(21C2−10C2)+(21C3−10C3)+(21C4−10C4)+...+(21C10−10C10) is
A
221−211
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B
221−210
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C
220−29
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D
220−210
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Solution
The correct option is D220−210 (21C1−10C1)+(21C2−10C2)+(21C3−10C3)+...+(21C10−10C10) =(21C1+21C2+⋯+21C10)−(10C1+10C2+⋯+10C10)=(21C1+21C2+⋯+21C10)−(210−1)=12(21C1+21C2+⋯21C21−1)−(210−1)=12(221−2)−(210−1)=220−1−210+1=220−210