5.082 If we expand all the brackets in the expression $(1+x)^{9}+(1+x)^{10}+\ldots$ $\ldots+(1+x)^{14}$ and combine like terms, we will get some polynomial. Determine the coefficient of $x^{9}$ in this polynomial without expanding the brackets.
Solution. The coefficient of $x^{9}$ is $A=C_{9}^{9}+C_{10}^{9}+C_{11}^{9}+C_{12}^{9}+C_{13}^{9}+C_{14}^{9}$. According to the condition of problem 5.076,

\[
\begin{aligned}
& A+C_{10}^{10}=1+C_{10}^{9}+C_{11}^{9}+\ldots+C_{14}^{9}+C_{10}^{10}=1+\left(C_{10}^{9}+C_{10}^{10}\right)+C_{11}^{9}+\ldots+C_{14}^{9}= \\
& =1+\left(C_{11}^{10}+C_{11}^{9}\right)+C_{12}^{9}+C_{13}^{9}+C_{14}^{9}=1+\left(C_{12}^{10}+C_{12}^{9}\right)+C_{13}^{9}+C_{14}^{9}=
\end{aligned}
\]

$=1+\left(C_{13}^{10}+C_{13}^{9}\right)+C_{14}^{9}=1+C_{14}^{10}+C_{14}^{9}=1+C_{15}^{10}$.

Therefore, $A=1+C_{15}^{10}-C_{10}^{10}=C_{15}^{10}=3003$.

Answer: 3003.