The value of 20C0+20C1+20C2+20C3+20C4+20C12+20C13+20C14+20C15 is
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a
219− 20C10+20C92
b
219− 20C10+2×20C92
c
219− 20C102
d
none of these
answer is B.
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Detailed Solution
Given series is 20C0+20C1+20C2+⋯+20C8 =122⋅20C0+220C1+⋯2⋅20C8 =12 20C0+20C1+⋯+20C8+20C9+20C10+20C11+…+20C20− 20C9+20C10+20C11 =12220−2⋅20C9−20C10 =219−2⋅20C9+20C102