Coefficient of x^{13} in expansion of (1 -x)^5(1 + x + x^2 + x^3)^4 is :-

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7.Binomial Theorem

normal

The coefficient of $x^{13}$ in the expansion of $(1 -x)^5(1 + x + x^2 + x^3)^4$ is :-

A

$-4$

B

$0$

C

$4$

D

none of these

Solution

$(1-x)^{5} \times(1+x)^{4}\left(1+x^{2}\right)^{4}$

$=(1-x)(1-x)^{4}(1+x)^{4} \times\left(1+x^{2}\right)^{4}$

$=(1-\mathrm{x})\left(1-\mathrm{x}^{4}\right)^{4}$

$=\left(1-x^{4}\right)^{4}-x\left(1-x^{4}\right)^{4}$

Coff. of ${x^{13}}0 + {\,^4}{C_3} = 4$

Standard 11

Mathematics

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