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Q.
The coefficient of $x^4$ in the expansion of $\left(1-x-x^2+x^3\right)^6$ is
TS EAMCET 2021
Solution:
$ \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 $
$=15-240+990-1320+495=-60$