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Q.

C0   30C10   30-C1   30C11   30+C2   30·C12   30+C20   30C30   30=

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a

C10   31

b

C15   30

c

C20   60

d

C10   30

answer is A.

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Detailed Solution

 (1-x)30=C0-C1x+C2x2++C30x30 1+1x30=C0+C11x+C21x2+.+C301x30 where Cr=Cr   30Now (1-x)301+1x30=C0-C1x++C30x30C0+C11x++C301x30C0C10-C1C11+C2C12+C20·C30= Coefficient of 1x10 in (1-x)30x+1x30= Coefficient of x20 in 1-x230=C10   30.

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C0   30C10   30-C1   30C11   30+C2   30·C12   30…+C20   30C30   30=