1.21 A children's abacus has ten beads on each of the three rods (as shown in the figure). The beads on each rod are to be divided into two parts (no part can be empty). Now, we want the product of the three numbers of beads on the left side to equal the product of the three numbers of beads on the right side. How many ways are there to distribute the beads?
（Beijing, China High School Grade 1 Mathematics Competition, 1987）
[Solution] (1) If all three compartments are evenly divided into two 5s, it is one way that meets the requirement.
$$
5 \cdot 5 \cdot 5=5 \cdot 5 \cdot 5 .
$$
(2) If two compartments are evenly divided into $(5,5)$, then the third compartment must also be divided into $(5,5)$, so it is still the same way as in (1).
(3) If one compartment is divided into $(5,5)$, and the other two compartments are not evenly divided. Let these two compartments be $r$ and $s (r \neq 5, s \neq 5)$, then the other side must be $10-r, 10-s$. At this time, when $r+s=10$, i.e.,
$$
r=10-s, \quad s=10-r \text {, }
$$

it forms a way that meets the requirement, because at this time
$$
5 \cdot r \cdot s=5(10-r)(10-s) .
$$

There are 8 ways to choose $r$: $1,2,3,4,6,7,8,9$. Thus, there are 8 pairs of $(r, s)$, and each pair of $5, r, s$ can be arranged on one side, so there are $3 \cdot 8=24$ ways to distribute the beads.
(4) Below, we prove that if there is no $(5,5)$ in the bead distribution, the requirement of the problem cannot be achieved. Suppose the number of beads in the three compartments from top to bottom on the left side are $a, b, c$ respectively. According to the problem, we have
$$
\begin{aligned}
a b c & =(10-a)(10-b)(10-c) \\
& =1000-100(a+b+c)+10(a b+b c+c a)-a b c, \\
a b c & =500-50(a+b+c)+5(a b+b c+c a) .
\end{aligned}
$$

Since $a, b, c \in\{1,2,3,4,5,6,7,8,9\}$, and every term on the right side of the above equation is a multiple of 5, at least one of $a, b, c$ must be 5. Therefore, there must be a $(5,5)$ in the bead distribution.
From the above, the total number of bead distribution methods that meet the requirement is $24+1=25$.