Question

Find the coefficient of $x^7$ in the expansion ${(1-x-x^2+x^3)}^6$

Answer 1

${(1-x-x^2+x^3)}^6$

$={\left((1-x)(1-x^2)\right)}^6$

$={(1-x)}^{12}.{(1+x)}^6$

Coefficient of $x^n$ in ${(1+x)}^6{=}^6{C}_n$

Coefficient of $x^n$ in ${(1-x)}^{12}={(-1)}^n\left({}^{12}{C}_n\right)$

Coefficient of $x^7$ in this expansion$=$

Coefficient of $x$ in ${(1-x)}^{12}.$Coefficient of $x^6$ in ${(1+x)}^6+$Coefficient of $x^2$ in ${(1-x)}^{12}.$Coefficient of $x^5$ in ${(1+x)}^6+$Coefficient of $x^3$ in ${(1-x)}^{12}.$Coefficient of $x^4$ in ${(1+x)}^6+$Coefficient of $x^4$ in ${(1-x)}^{12}.$Coefficient of $x^3$ in ${(1+x)}^6+$Coefficient of $x^5$ in ${(1-x)}^{12}.$Coefficient of $x$ in ${(1+x)}^6+$Coefficient of $x^7$ in ${(1-x)}^{12}.$Coefficient of $x^0$ in ${(1+x)}^6$

$=-\left({}^{12}{C}_1{.}^6{C}_6\right)+\left({}^{12}{C}_2{.}^6{C}_5\right)-\left({}^{12}{C}_3{.}^6{C}_4\right)+\left({}^{12}{C}_4{.}^6{C}_3\right)-\left({}^{12}{C}_5{.}^6{C}_2\right)+\left({}^{12}{C}_6{.}^6{C}_1\right)-\left({}^{12}{C}_7{.}^6{C}_0\right)$

$=-12+(66\times 6)-(220\times 15)+(495\times 20)-(792\times 15)-(792\times 15)+(132\times 42)-792$(on simplification)

$=-144$