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Q.

The value of 20C0+20C1+20C2+20C3+20C4+20C12+20C13+20C14+20C15  equals to

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a

21920C102

b

21920C10220C92

c

219(20C0+20C9)2

d

219(20C10+2×20C9)2

answer is B.

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Detailed Solution

20C0+20C1+20C2+20C3+20C4+20C12+20C13+20C14+20C15

=20C0+20C1+20C2+20C3+20C4+20C5+20C6+20C7+20C8

=12(2(20C0+20C1++20C8))

=12(2.20C0+2.20C1++2.20C8)

=12(20C0+20C20+20C1+20C19+20C2+20C18+)+(20C9+20C10+20C11)(20C9+20C10+20C11)

=12(2202.20C920C10)

=219(2.20C9+20C10)2

=21920C10220C92

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