We know $$(1+x)^n={ }^n \mathrm{C}_0+{ }^n \mathrm{C}_1 x+{ }^n \mathrm{C}_2 x^2+{ }^n C_3 x^3+\ldots .+{ }^n C_n x^n$$.

$$\begin{aligned} & \text { Given, }\left(1+x^2\right)^4\left(1+x^3\right)^7\left(1+x^4\right)^{12} \\ & \Rightarrow\left({ }^4 \mathrm{C}_0+{ }^4 \mathrm{C}_1 x^2+{ }^4 \mathrm{C}_2 x^4+{ }^4 \mathrm{C}_3 x^6+{ }^4 \mathrm{C}_4 x^8\right) \\ & \quad \times\left({ }^7 \mathrm{C}_0+{ }^7 \mathrm{C}_1 x^3+{ }^7 \mathrm{C}_2 x^6+{ }^7 \mathrm{C}_3 x^9+{ }^7 \mathrm{C}_4 x^{12}+\cdots\right) \\ & \quad \times\left({ }^{12} \mathrm{C}_0+{ }^{12} \mathrm{C}_1 x^4+{ }^{12} \mathrm{C}_2 x^8+{ }^{12} \mathrm{C}_3 x^{12}+\ldots .\right) \end{aligned}$$

$$\Rightarrow$$ The coefficient of

$$\begin{aligned} & x^{11}={ }^4 \mathrm{C}_0 \cdot{ }^7 \mathrm{C}_1 \cdot{ }^{12} \mathrm{C}_2+{ }^4 \mathrm{C}_1 \cdot{ }^7 \mathrm{C}_3 \cdot{ }^{12} \mathrm{C}_0 \\ & +{ }^4 \mathrm{C}_2 \cdot{ }^7 \mathrm{C}_1 \cdot{ }^{12} \mathrm{C}_1+{ }^4 \mathrm{C}_4 \cdot{ }^7 \mathrm{C}_1 \cdot{ }^{12} \mathrm{C}_0 \end{aligned}$$

$$\Rightarrow$$ The coefficient of $$x^{11}=462+140+504+7$$

$$\Rightarrow$$ The coefficient of $$x^{11}=1113$$

Hint: (i) Recall

$$\begin{aligned} & (1+x)^n={ }^n \mathrm{C}_0+{ }^n \mathrm{C}_1 x+{ }^n \mathrm{C}_2 x^2 \\ & +{ }^n \mathrm{C}_3 x^3+{ }^n \mathrm{C}_4 x^4+\ldots \ldots+{ }^n \mathrm{C}_n x^n . \end{aligned}$$

(ii) $${ }^n \mathrm{C}_r=\frac{n!}{r!(n-r)!}$$