The coefficient of $x^{3}$ in the expansion of ${(1 - x + x^{2})}^{5}$ is

10

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- 20

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- 50

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- 30

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Solution

The correct option is D $- 30$
$(1 - x + x^{2})^{5}$
Writing the general term
$T_{r + 1} =^{5} C_{r} (x^{2} - x)^{r}$
$= (- 1)^{k}^{5} C_{r}^{r} C_{k} x^{2 r - 2 k} x^{k}$
For coefficient of $x^{3}$
$2 r - k = 3$
Let $r = 2$
Hence $1 = k$
$k = 1$
Substituting, we get he coefficient as
$-^{5} C_{2} .^{2} C_{1}$
$= - 10 (2)$
$= - 20$ ...(i)
Let $r = 3$
Hence $k = 3$
The coefficient being
$(- 1)^{3}^{5} C_{3}^{3} C_{3}$
$= - 10$ ...(iii)
Hence the coefficient of $x^{3}$ will be
sum of i and ii
that is $- 10 + (- 20)$
$= - 30$

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