$4 \cdot 220$ There is a group of children, the sum of their ages is 50 years, the oldest is 13 years old, and one is 10 years old; except for the 10-year-old child, the ages of the remaining children form an arithmetic sequence. How many children are there? How old is each child?
[Solution] Let the age of the oldest child be $a$, i.e., $a=13$.
Suppose there are $b+1$ children excluding the 10-year-old child, and let the common difference of their ages be $d$. Then the ages of these $b+1$ children are
$$a, a-d, a-2 d, \cdots, a-b d$$

Thus, $a+(a-d)+(a-2 d)+\cdots+(a-b d)$
$$=(b+1) a-\frac{b(b+1)}{2} d=50-10=40,$$

which means $(b+1)(2 a-b d)=80$.
It is clear that $(b+1)$ divides 80, but $2 a-b d=a+(a-b d)>13$, so $b+1<\frac{80}{26}$, considering both aspects, we know that $b+1$ can only be 4 or 5.

When $b+1=4$, $4(26-3 d)=80$, hence $d=2$. Substituting the values of $b$ and $d$ into (1) gives: $13,11,10,9,7$. Therefore, there are $4+1=5$ children, and their ages are $13,11,10,9,7$.
When $b+1=5$, $5(26-4 d)=80$, in this case $d$ is not an integer, so this situation is impossible. Therefore, the solution is unique.