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# The coefficient of $x^7$ in the expansion $(1-x-x^2+x^3)^6$ is ?

$\begin{array}{1 1} -144 \\ -132 \\ 132 \\ 144 \end{array}$

Can you answer this question?

$(1-x-x^2+x^3)^6=\big((1-x)(1-x^2)\big)^6$
$=(1-x)^{12}.(1+x)^6$
Coefficient of $x^n$ in $(1+x)^{6}$ = $^{6}C_n$
Coefficient of $x^n$ in $(1-x)^{12}$ = $(-1)^n$.$^{12}C_n$
Coeff. of $x^7$ in this expansion =
coeff. of $x$ in $(1-x)^{12}$.coeff of $x^6$ in $(1+x)^6+$
coeff. of $x^2$ in $(1-x)^{12}$.coeff of $x^5$ in $(1+x)^6+$
coeff. of $x^3$ in $(1-x)^{12}$.coeff of $x^4$ in $(1+x)^6+$
coeff. of $x^4$ in $(1-x)^{12}$.coeff of $x^3$ in $(1+x)^6+$
coeff. of $x^5$ in $(1-x)^{12}$.coeff of $x^2$ in $(1+x)^6+$
coeff. of $x^6$ in $(1-x)^{12}$.coeff of $x$ in $(1+x)^6+$
coeff. of $x^7$ in $(1-x)^{12}$.coeff of $x^0$ in $(1+x)^6+$
=
$-(^{12}C_1.^6C_6)+(^{12}C_2.^6C_5)-(^{12}C_3.^6C_4)+(^{12}C_4.^6C_3)$
$-(^{12}C_5.^6C_2)+(^{12}C_6.^6C_1)-(^{12}C_7.^6C_0)$
$=-12+(66\times 6)-(220\times 15)+(495\times 20)-(792\times 15)+(132\times 42)-792$
$=-144$

answered Sep 17, 2013