Q.

The value of  21C1−10C1+ 21C2−10C2 + 21C3−10C3+ 21C4−10C4+…++ 21C10−10C10 is

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a

221−211

b

221−210

c

220−29

d

220−210

answer is D.

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Detailed Solution

We have, 21C1−10C1+ 21C2−10C2+ 21C3−10C3+ 21C4−10C4+…+ 21C10−10C10 21C1+21C2+21C3+21C4+…+21C10− 10C1+10C2+10C3+10C4+…+10C10=12221C1+221C2+221C3+221C4+…+221C10− 10C1+10C2+10C3+10C4+…+10C10= 21C1+21C2+21C3+21C4+…+21C10− 10C1+10C2+10C3+10C4+…+10C10=12221C1+221C2+221C3+221C4+…+221C10=12 21C1+21C20+ 21C2+21C19+ 21C3+21C18+…+ 21C10+21C11− 10C1+10C2=12221− 21C0+21C21−210−10C0=12221−2−210−1=220−210
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