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Permutations , combinations and binomial expansion — Statistics STD 11 Commerce — Question

Gujarat BoardEnglish MediumSTD 11 CommerceStatisticsPermutations , combinations and binomial expansion2 Marks
Question
Expand $\left(x-\frac{2}{x}\right)^{5}$
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Answer

$\left(x-\frac{2}{x}\right)^{5}$
$={ }^{5} \mathrm{C}_{0}(x)^{5}\left(\frac{2}{x}\right)^{0}-{ }^{5} \mathrm{C}_{1}(x)^{4}\left(\frac{2}{x}\right)^{1}+{ }^{5} \mathrm{C}_{2}(x)^{3}\left(\frac{2}{x}\right)^{2}-{ }^{5} \mathrm{C}_{3}(x)^{2}\left(\frac{2}{x}\right)^{3}+{ }^{5} \mathrm{C}_{4}(x)^{1}\left(\frac{2}{x}\right)^{4}-{ }^{5} \mathrm{C}_{5}(x)^{0}\left(\frac{2}{x}\right)^{5}$
$=x^{5}-5(x)^{4}\left(\frac{2}{x}\right)+10(x)^{3}\left(\frac{4}{x^{2}}\right)-10\left(x^{2}\right)\left(\frac{8}{x^{3}}\right)+5(x)\left(\frac{16}{x^{4}}\right)-\frac{32}{x^{5}}$
$=x^{5}-10 x^{3}+40 x-\frac{80}{x}+\frac{80}{x^{3}}-\frac{32}{x^{5}}$

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