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  4. The coefficient of x7 in the expansion of (1-x-x3+x4)8 is

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Q. The coefficient of $x^{7}$ in the expansion of $\left(1-x-x^{3}+x^{4}\right)^{8}$ is____

Binomial Theorem

Solution:

$\left(1-x-x^{3}+x^{4}\right)^{8}=(1-x)^{8}\left(1-x^{3}\right)^{8}$
$=\left(1-{ }^{8} C_{1} x+{ }^{8} C_{2} x^{2}-{ }^{8} C_{2} x^{3}+\ldots\right)$
$\left(1-{ }^{8} C_{1} x^{3}+{ }^{8} C_{2} x^{6}-{ }^{8} C_{3} x^{9}+\ldots\right)$
$\therefore $ Coefficient of $x^{7}$
$=-{ }^{8} C_{7}-{ }^{8} C_{1} \times{ }^{8} C_{4}-{ }^{8} C_{1} \times{ }^{8} C_{2}$
$=-792$