10. (10 points) 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 among 8 children, with exactly all gifts being distributed (each child gets at least one gift). How many different selections are there? $\qquad$
【Analysis】36 identical gifts are packed 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 the number of gifts in the bags that are multiples of 8, and we can enumerate the cases accordingly.

【Solution】Solution: If each person gets 1 gift: $8=<8=1+7=2+6=3+5=1+2+5=1+3+4,6$ ways; If each person gets 2 gifts: $16=1+7+8=2+6+8=3+5+8=3+6+7=4+5+7=1+2+5+8=1+2+6+7$ $=1+3+4+8=1+3+5+7=1+4+5+6=2+3+4+7=2+3+5+6$ $=1+2+3+4+6$, a total of 13 ways;

If each person gets 3 gifts, decomposing 24 is the same as decomposing $36-24=12$.
$$
\begin{array}{l}
12=4+8=5+7 \\
=1+3+8=1+4+7=1+5+6=2+3+7=2+4+6=3+4+5
\end{array}
$$
$=1+2+3+6=1+2+4+5$, a total of 10 ways;
If each person gets 4 gifts, similarly decomposing $36-32=4$ $4=4=1+3$, a total of 2 ways;

Therefore, there are $6+13+10+2=31$ different ways to choose.
The answer is: 31.