A) 130
B) 120
C) 128
D) 125
E) 115
Correct Answer: B
Solution :
\[{{(1+x+{{x}^{2}}+{{x}^{3}})}^{6}}\] \[={{(1+x)}^{6}}{{(1+{{x}^{2}})}^{6}}\] \[={{(}^{6}}{{C}_{0}}{{+}^{6}}{{C}_{1}}x{{+}^{6}}{{C}_{2}}{{x}^{2}}{{+}^{6}}{{C}_{3}}{{x}^{3}}{{+}^{6}}{{C}_{4}}{{x}^{4}}\] \[{{+}^{6}}{{C}_{5}}{{x}^{5}}{{+}^{6}}{{C}_{6}}{{x}^{6}})\] \[\times {{(}^{6}}{{C}_{0}}{{+}^{6}}{{C}_{1}}{{x}^{2}}{{+}^{6}}{{C}_{2}}{{x}^{4}}{{+}^{6}}{{C}_{3}}{{x}^{6}}{{+}^{6}}{{C}_{4}}{{x}^{8}}\] \[{{+}^{6}}{{C}_{5}}{{x}^{10}}{{+}^{6}}{{C}_{6}}{{x}^{12}})\] \[\therefore \]Coefficient of\[{{x}^{14}}\]in\[{{(1+x+{{x}^{2}}+{{x}^{3}})}^{6}}\] \[{{=}^{6}}{{C}_{2}}{{.}^{6}}{{C}_{6}}{{+}^{6}}{{C}_{4}}{{.}^{6}}{{C}_{5}}{{+}^{6}}{{C}_{6}}{{.}^{6}}{{C}_{4}}\] \[=\frac{6!}{2!4!}.\frac{6!}{0!6!}+\frac{6!}{4!2!}.\frac{6!}{5!1!}+\frac{6!}{0!6!}.\frac{6!}{2!4!}\] \[=15+90+15=120\]
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