Question

The coefficient of $x^4$ in the expansion of ${\left(1-x-x^2+x^3\right)}^6$ is

• A. 120
• B. 15
• C. $-75$
• D. $-60$

Answer 1

Answer: (D)

${\left(1-x-x^2+x^3\right)}^6={\left[(1-x)\left(1-x^2\right)\right]}^6$
$=(1-x{)}^{12}(1+x{)}^6$
In the expansion of $(1-x{)}^{12}$, coefficient are of the form $(-1{)}^{r^{12}}{C}_r$ and in $(1+x{)}^{12}$, coefficient are of the form ${}^6{C}_r$
Coefficient of $x^4$ in expansion of ${\left(1-x-x^2+x^3\right)}^6$
$={}^{12}{C}_0\times {}^6{C}_4-{}^{12}{C}_1\times {}^6{C}_3+{}^{12}{C}_2\times {}^6{C}_2$
$-{}^{12}{C}_3\times {}^6{C}_1+{}^{12}{C}_4\times {}^6{C}_0$
$=15-240+990-1320+495=-60$