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a
12C10 20
b
0
c
C10 20
d
-C10 20
answer is A.
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Detailed Solution
We know that(1−1)20=20C0−20C1+20C2−20C3+…+20C10−20C11+20C12−⋯+20C20=02 20C0−20C1+20C2−20C3+⋯−20C9+20C10=0 ∵20C20=20C0,20C19=20C1, etc. ⇒20C0−20C1+20C2−20C3+⋯−20C9+20C10=−12C10 20+20C10=12C10 20