Q.

 11C109C1+11C99C2+.+11C29C9==

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a

Number of different ways of exchanging 11 books of A with the 9 books of B

b

 20C91

c

 20C11

d

 20C8

answer is C, D.

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Detailed Solution

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(1+x)11=11C0+11C1x+11C2x2++11C11x11(I)

(1+x)9=9C0+9C1x+9C2x2++9C9x9-(II)

On multiply (I) & (II) and compare coefficient  of  x11on both sides and put  x=1
 20C11=11C119C0+11C109C1+..+11C29C9

  20C91=11C109C1+.+11C29C9

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 11C10⋅9C1+11C9⋅9C2+…….+11C2⋅9C9==