Correct answer is : 70
\(x_1+x_2+x_3=14 \text {, where } 1 \leq x_1 \leq 9,0 \leq x_2, x_3 \leq 9\)
\(\text { Coefficient of } x^{14} \text { in } \Rightarrow\left(x^1+x^2+\ldots x^9\right)\left(x^0+x^1+x^2+\ldots+x^9\right)^2 \)
\( \Rightarrow x\left(1+x+\ldots x^8\right)\left(1+x+x^2+\ldots x^9\right)^2\)
\(\Rightarrow x\left(\frac{1-x^9}{1-x}\right) \cdot\left(\frac{1-x^{10}}{1-x}\right)^2\)
\( \Rightarrow x \cdot\left(1-x^9\right)\left(1-x^{10}\right)^2 \cdot(1-x)^{-3}\)
\( \Rightarrow x \cdot\left(1-x^9\right)\left(1-2 x^{10}+x^{20}\right)(1-x)^{-3} \)
\( \Rightarrow x \cdot\left(1-2 x^{10}+x^{20}-x^9+2 x^{19}-x^{29}\right)(1-x)^{-3}\)
\( \Rightarrow x\left(1-2 x^{10}-x^9+\ldots .\right)(1-x)^{-3} \quad\{\text { ignoring higher power\} } \)
\(\text { No. of integer } \Rightarrow{ }^{3+13-1} C_{13}-2\left({ }^{3+3-1} C_3\right)-\left({ }^{3+4-1} C_4\right)\)
\(\Rightarrow{ }^{15} C_{13}-2\left({ }^5 C_3\right)-{ }^6 C_4 \Rightarrow 105-20-15 \Rightarrow 70\)
