Find the coeff of $x^{5}$ in the expansion of the product $(x^{2} - x - 2)^{5}$

- 83

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- 82

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- 71

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- 81

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Solution

The correct option is D $- 81$
General term of ${(x^{2} - x - 2)}^{5}$ is
$\frac{5!}{r_{1} {! r}_{2}! r_{3}!} {(x^{2})}^{r_{1}} . {(- x)}^{r_{2}} . {(- 2)}^{r_{3}} r_{1} + r_{2} + r_{3} = 5 = \frac{5! . {(- 1)}^{r_{2}} . {(- 2)}^{r_{3}}}{r_{1} {! r}_{2}! r_{3}!} . {(x)}^{{2 r}_{1} + r_{2}} {2 r}_{1} + r_{2} = 5$
So, the possible combinations can be $r_{1} {= 0 r}_{2} = 5 r_{3} = 0 r_{1} {= 1 r}_{2} = 3 r_{3} = 1 r_{1} {= 2 r}_{2} = 1 r_{3} = 2$
Taking these values, coefficient of $x^{5} = \frac{5! . (- 1)}{5!} + \frac{5! . (- 1) (- 2)}{3!} + \frac{5! . (- 1) .4}{2! 2!} = - 1 + 40 - 120 = - 81$

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