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Question

For r=0,1,...,10, let Ar, Br, and Cr denote, respctively, the coefficient of xr in the expansions of (1+x)10, (1+x)20 and (1+x)30. Then 10r=1Ar(B10BrC10Ar)is equal to

A
B10C10
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B
A10(B210C10A10)
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C
\N
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D
C10B10
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Solution

The correct option is D C10B10
Clearly Ar=10Cr,Br=20Cr,Cr=30Cr,Now 10r=110Cr(20C10 20Cr 30C10 10Cr)=20C1010r=1 10Cr 20Cr 30C1010r=1 10Cr×10Cr=20C10(10C1 20C1+ 10C2 20C2+...+10C10 20C10)=30C10(10C1×10C1+10C2×10C2+...+10C10 10C10)...(1)
Now expanding (1+x)10 and (1+x)20 by binomial theorem and comparing the coefficients of x20 in their product, on both sides, we get
10C0 20C0+10C1 20C1+10C2 20C2+...+10C10 20C10
= coeff of x20 in (1+x)30=30C20=30C1010C1 20C1+10C2 20C2+...+10C10 20C10=30C101
Again expending (1+x10) and (x+1)10 by binomial theorem and comparing the coefficients of x10 in their product on both sides, we get
(10C0)2(10C1)2+(10C2)2+...+(10C10)2
= coeff of x10 in (1+x)20=20C10(10C1)2+(10C2)2+...+(10C10)2=20C101
Substituting these values in equation (1), we get
20C10(30C101)30C10(20C101)=30C1020C10=C10B10

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