Find the coefficient of
$x^{8}$ in the expansion of $(1 - 2 x + 3 x^{2} - 4 x^{3})^{4}$ .

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Solution

General term of expansion is $\frac{5!}{a! b! c! d!} (- 2)^{b} 3^{c} (- 4)^{d} x^{b + 2 c + 3 d}$ , where $a + b + c + d = 4$

We want to find the coefficient of $x^{8}$ , therefore $b + 2 c + 3 d = 8$ . This is possible for,

$a = 0, b = 0, c = 4, d = 0; a = 0, b = 1, c = 2, d = 1; a = 0, b = 2, c = 0, d = 2; a = 1, b = 0, c = 1, d = 2$

$∴$ the coefficient of $x^{4} = \frac{4!}{0! 0! 4! 0!} (- 2)^{0} 3^{4} (- 4)^{0} + \frac{4!}{0! 1! 2! 1!} (- 2)^{1} 3^{2} (- 4)^{1} + \frac{4!}{0! 2! 0! 2!} (- 2)^{2} 3^{0} (- 4)^{2} + \frac{4!}{1! 0! 1! 2!} (- 2)^{0} 3^{1} (- 4)^{2}$

$= 81 + 864 + 384 + 576 = 1905$

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