The coefficient of $x$ in the expansion of $(1 - x - x^{2} + x^{3})^{6}$ is ?

6

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- 6

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- 12

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12

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Solution

The correct option is C $- 12$
${(1 - x - x^{2} + x^{3})}^{6} = {((1 - x) (1 - x^{2}))}^{6} = {(1 - x)}^{12} . {(1 + x)}^{6}$

Coefficient of xn in ${(1 + x)}^{6} = 6 C_{n}$

Coefficient of xn in ${(1 - x)}^{12} = {(- 1)}^{n} .12 C_{n}$

Coeff.of{ x }^{ 7 }in this expansion $= - (12 C_{1} .6 C_{6}) + (12 C_{2} .6 C_{5}) - (12 C_{3} .6 C_{4}) + (12 C_{4} .6 C_{3})$

$= (12 C_{5} .6 C_{2}) + (12 C_{6} .6 C_{1}) - (12 C_{7} .6 C_{0})$

$= - 12 + (66 \times 6) - (220 \times 15) + (495 \times 20) - (792 \times 15) + (132 \times 42) - 792$

$= - 12$

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