Answer: 122
Clearly, the balls numbered $15,14, \ldots, 9,8$ must be placed in separate bins, so we number the bins $15,14, \ldots, 9,8$. Note that bins 15 and 14 may contain only one ball while all other bins must contain at least two balls. We have two cases to examine.
Case 1: Only one bin contains exactly one ball. Let $a_{i}$ denote the number of ways to place the balls numbered $1,2, \ldots, i-1$ into the bins numbered $15,14, \ldots, 15-i+1$. We can place either $i-1$ or $i-2$ into the bin numbered $15-i+1$. If we place $i-1$ in there, then there are $a_{i-1}$ ways to finish packing the rest. If we place $i-2$ in this bin, then $i-1$ must be placed in the bin numbered $15-i+2$, so there are $a_{i-2}$ ways to place the rest of the balls. Therefore $a_{i}=a_{i-1}+a_{i-2}$. Since $a_{1}=2$ and $a_{2}=3$, the sequence $\left\{a_{i}\right\}$ is the Fibonacci sequence, and $a_{7}=34$.
Case 2: Both bins 14 and 15 contain only one ball. A pair of balls from 1-7 must be put together to one of the bins numbered 8 through 13. This pair has sum at most 8 , so we can count for all the cases.
\begin{tabular}{c|c} 
Balls & Number of packings \\
\hline 1,2 & 16 \\
\hline 1,3 & 10 \\
\hline 1,4 & 12 \\
\hline 1,5 & 12 \\
\hline 1,6 & 10 \\
\hline 1,7 & 8 \\
\hline 2,3 & 6 \\
\hline 2,4 & 4 \\
\hline 2,5 & 4 \\
\hline 2,6 & 3 \\
\hline 3,4 & 2 \\
\hline 3,5 & 1
\end{tabular}

Therefore, there are 88 possibilities in this case, and the total number of possibilities is 122 .
14. $[8]$ Danny has a set of 15 pool balls, numbered $1,2, \ldots, 15$. In how many ways can he put the balls in 8 indistinguishable bins such that the sum of the numbers of the balls in each bin is 14,15 , or 16 ?