China Mathematical Competition

We may use each space between every two consecutive bars ($|$) to represent a school and each asterisk ($\ast$) to represent a volunteer, as seen in the following example; the first, second and third schools receive $4$, $18$ and $2$ volunteers, respectively. $$|\ast \ast \ast \ast |\ast \cdots \ast |\ast \ast |$$ Then the allocation problem may be regarded as a permutation-and-combination problem of $4$ bars and $24$ asterisks.

Since the two ends of the line must be occupied by a bar, respectively, there are $\left(\genfrac{}{}{0ex}{}{23}{2}\right)=253$ ways to insert the other $2$ bars into the $23$ spaces between the $24$ asterisks such that there is at least $1$ asterisk between every two consecutive bars, in which there are $31$ ways that at least two schools have the same number of volunteers. So the number of allocating ways satisfying the conditions is $253-31=222$.