10. Santa Claus has 36 identical gifts, which are placed in 8 bags. It is known that the number of gifts in the 8 bags is at least 1 and all different. Now, some bags are to be selected, and all the gifts in the selected bags are to be evenly distributed to 8 children, with exactly all gifts being distributed (each child gets at least one gift). Therefore, there are $\qquad$ different ways to choose.
Analysis: Number theory problem, integer partition.
To solve this problem, a bit of number sense is required. 36 identical gifts are to be placed in 8 bags, with each bag containing at least 1 gift and the number of gifts in each bag being different. Since $1+2+3+\cdots+8=(1+8) \times 8 \div 2=36$, it is clear that the 8 bags contain 1 to 8 gifts respectively. According to the problem, we need to select bags where the number of gifts is a multiple of 8. We can enumerate the cases accordingly.
If each person gets 1 gift: $8=8$
$$
\begin{array}{l}
=1+7=2+6=3+5 \\
=1+2+5=1+3+4, 6 \text{ ways; }
\end{array}
$$

If each person gets 2 gifts:
$$
\begin{aligned}
16 & =1+7+8=2+6+8=3+5+8=4+5+7 \\
& =1+2+5+8=1+2+6+7=1+3+4+8=1+3+5+7=1+4+5+6=2+3+4+7=2+3+5+6 \\
& =1+2+3+4+6, \text{ a total of } 13 \text{ ways; }
\end{aligned}
$$

If each person gets 3 gifts, partition 24, which is the same as partitioning $36-24=12$.
$$
\begin{aligned}
12 & =4+8=5+7 \\
& =1+3+8=1+4+7=1+5+6=2+3+7=2+4+6=3+4+5 \\
& =1+2+3+6=1+2+4+5, \quad \text{ a total of } 10 \text{ ways; }
\end{aligned}
$$

If each person gets 4 gifts, similarly partition $36-32=4$
$$
4=4=1+3 \text{, a total of } 2 \text{ ways; }
$$

Therefore, there are a total of $6+13+10+2=31$ different ways to choose.
The key to this problem is to enumerate in an orderly manner, without repetition or omission!