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Q.

If a is the coefficient of  x5 in (1x+x2x3)5,b is the coefficient of  x7in (1xx2+x3)5 and c is the coefficient of   in (1xx2+x3)6, then the descending order  of a, b, c is 

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a

a,b,c

b

b,a,c

c

c,a,b

d

a,c,b

answer is B.

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Detailed Solution

(1x+x2x3)5

=(1x)+x2(1x)5

=(1x)5(1+x2)5

=(5C05C1x+5C2x25C3x3+5C4x45C5x5) ×(5C0+5C1x +5C2x4+)

Given a=x5 coefficient

=5C1.5C25C3.5C15C5.5C0

=[5×10+5×10+1]

=101(1xx2+x3)5((1x)x2(1x))5(1x)5(1x2)5(1x)10(1+x)5b=coefficient  of  x7=[10C010C1x+10C2x2+][5C0+5C1x+5C2x2+]

=10C2.5C5+(10C3)5C4+10C45C310C55C2+10C65C110C7.5C0

=45600+21002520+1050120

=45

(1xx2+x3)6(1xx2+(1x))6(1x)6+(1x2)6(1x)12+(1+x)6C=coefficient.  of  x7  in  (1xx2+x3)6=(1x)12(1+x)6

(1x)12(1+x)6=(12C012C1x+12C2x2+)(6C0+6C1x+)x7coefficient=12C1.6C6+12C26C512C3.6C4+12C46C312C56C2+12C66C112C76C0

C=(12)+(66)(6)3300+990011880+3128792=(396+9900+3128)(12+3300++11880+792)=13,42415,984C=2560

Descending order is b, a, c

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