If xm occurs in the expansion of (x+1x2)2n, then the coefficient of xm is
A
(2n)!(m)!(2n−m)
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B
(2n)!3!3!(2n−m)!
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C
(2n)!(2n−m3)!(4n+m3)!
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D
(2n)!(2n−m3)!(3n+m2)!
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Solution
The correct option is C(2n)!(2n−m3)!(4n+m3)! Tr+1=2nCrx2n−r(1x2)r=2nCrx2n−3r This contains xm, If 2n−3r=m ⇒r=2n−m3 Hence, coefficient of xm=2nCr =2n!(2n−r)!r!