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Question

Find the coefficient of $${ x }^{ 4 }$$ in the expansion of $${ \left( 2-x+{ 3x }^{ 2 } \right)  }^{ 6 }$$


A
3661
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B
1830
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C
3300
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D
none of the above
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Solution

The correct option is D none of the above
We want the coefficient of $$x^{4}$$ in $$(2x^{0}-1x+3x^{2})^{6}$$ will be in the case $$(x^{0})^{3}(x^{1})^{1}(x^{2})^{2}$$ , $$(x^{0})^{2}(x^{1})^{4}(x^{2})^{0}$$ and $$(x^{0})^{4}(x^{1})^{0}(x^{2})^{2}$$
Coefficient of $$x^{4}$$ in:
$$(x^{0})^{3}(x^{1})^{1}(x^{2})^{2}=\displaystyle \frac{6!(2)^{3}(-1)^{1}(3)^{2}}{3!1!2!}=-4320$$
$$(x^{0})^{2}(x^{1})^{4}(x^{2})^{0}=\displaystyle \frac{6!(2)^{2}(-1)^{4}(3)^{0}}{2!4!0!}=60$$ 
$$(x^{0})^{4}(x^{1})^{0}(x^{2})^{2}=\displaystyle \frac{6!(2)^{4}(-1)^{0}(3)^{2}}{4!0!2!}=2160$$ 
So, total will be $$2160+60-4320=-2100$$
i.e., none of the above.

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