32. Determine the coefficient of $x^{29}$ in the expansion of $\left(1+x^{5}+x^{7}+x^{9}\right)^{16}$.
32. Answer: 65520
Note that
$$
\left(1+x^{5}+x^{7}+x^{9}\right)^{16}=\sum_{i=0}^{16}\binom{16}{i} x^{53}\left(1+x^{2}+x^{4}\right)^{4}
$$

It is clear that if $i>5$ or $i<4$, then the coefficient of $x^{29}$ in the expansion of $x^{51}\left(1+x^{2}+x^{4}\right)^{\prime}$ is 0 . Note also that if $i$ is even, then the coefficient of $x^{29}$ in the expansion of $x^{5 i}\left(1+x^{2}+x^{4}\right)^{1}$ is also 0 . Thus we only need to determine the coefficient of $x^{29}$ in the expansion of $\binom{16}{i} x^{51}\left(1+x^{2}+x^{4}\right)^{\prime}$ for $i=5$.
When $i=5$, we have
$$
\begin{aligned}
\binom{16}{i} x^{5 i}\left(1+x^{2}+x^{4}\right)^{\prime} & =\binom{16}{5} x^{25}\left(1+x^{2}+x^{4}\right)^{5} \\
& =\binom{16}{5} x^{25} \sum_{j=0}^{5}\binom{5}{j}\left(x^{2}+x^{4}\right)^{\prime} \\
& =\binom{16}{5} x^{25} \sum_{j=0}^{5}\binom{5}{j} x^{2 j}\left(1+x^{2}\right)^{j} .
\end{aligned}
$$

It is clear that the coefficient of $x^{4}$ in the expansion of $\sum_{j=0}^{5}\binom{5}{j} x^{2 /}\left(1+x^{2}\right)^{j}$ is $\binom{5}{1}+\binom{5}{2}=15$. Hence the answer is $\binom{16}{5} \times 15=65520$.