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Coefficient of $x^{11}$ in the expansion of $\left(1+x^2\right)^4\left(1+x^3\right)^7\left(1+x^4\right)^{12}$ is
$1051$
$1106$
$1113$
$1120$
Solution
Coefficent of $x^{11} \equiv \frac{\left(1+x^2\right)^4\left(1+x^3\right)^7\left(1+x^4\right)^{12}\left(1-x^2\right)^4}{\left(1-x^2\right)^4}$
Coefficent of $x^{11} \equiv\left(1-x^8\right)^4\left(1+x^4\right)^8\left(1+x^3\right)^7\left(1-x^2\right)^{-4}$
$=\left(1-4 x^8\right)\left(1+x^4\right)^8\left(7 x^3+35 x^9\right)\left(1-x^2\right)^{-4} $
$=\left(7 x^3+35 x^9-28 x^{11}\right)\left(1+x^4\right)^8\left(1-x^2\right)^{-4}$
Coefficent of $x^8=\left(7 x+35 x^6-28 x^8\right)\left(1+8 x^4+28 x^8\right)\left(1-x^2\right)^{-4}$
$=\left(7+35 x^6-28 x^3+56 x^4+196 x^8\right)\left(1-x^2\right)^{-4}$
Coefficent of $t ^4 \equiv\left(7+56 t ^2+35 t ^3+168 t ^4\right)(1- t )^{-4}$
$=7 \cdot{ }^7 C _3+56 \cdot{ }^5 C _3+35 \cdot{ }^4 C _3+168 $
$=245+700+168=1113 .$
$Alterantive :$
$2 x+3 y+4 z=11 $
$(x, y, z)=(0,1,2){ }^4 C_0 \times{ }^7 C_1 \times{ }^{12} C_2 $
$(1,3,0)^4 C_1 \times{ }^7 C_3 $
$(2,1,1)^4 C_2 \times{ }^7 C_1 \times{ }^{12} C_1 $
$(4,1,0)^7 C_1 $
$\text { coefficient of } x^{11} =66 \times 7+35 \times 4+42 \times 12+7 $
$\quad=1113$ Ans.