A) 15
B) 10
C) 14
D) 12
Correct Answer: A
Solution :
\[{{\left( \frac{1-{{t}^{6}}}{1-t} \right)}^{3}}\] \[={{\left( 1-{{t}^{6}} \right)}^{3}}{{\left( 1-t \right)}^{-3}}\] \[=\text{ }\left( 1{{-}^{3}}{{C}_{1}}{{t}^{6}}{{+}^{3}}{{C}_{2}}{{t}^{12}}{{-}^{3}}{{C}_{3}}{{t}^{18}} \right)\times {{\left( 1-t \right)}^{-3}}\] coefficient of \[{{t}^{4}}\] is \[1\times \]coefficient of \[{{t}^{4}}\] in\[{{\left( 1-t \right)}^{-}}^{3}=1\times {}^{3+4-1}{{C}_{4}}\] (By multinomial theorem) \[{{=}^{\text{6}}}{{\text{C}}_{\text{4}}}=15\]
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