Willy Wonka has $n$ distinguishable pieces of candy that he wants to split into groups. If the number of ways for him to do this is $p(n)$, then we have 
        \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}\hline
            $n$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ \\\hline
            $p(n)$ & $1$ & $2$ & $5$ & $15$ & $52$ & $203$ & $877$ & $4140$ & $21147$ & $115975$ \\\hline
        \end{tabular}
    Define a [i]splitting[/i] of the $n$ distinguishable pieces of candy to be a way of splitting them into groups. If Willy Wonka has $8$ candies, what is the sum of the number of groups over all splittings he can use?

[i]2020 CCA Math Bonanza Lightning Round #3.4[/i]
1. Define \( q(n, k) \) as the number of ways to split \( n \) distinguishable pieces of candy into \( k \) groups.
2. The total number of ways to split \( n \) pieces of candy into any number of groups is given by \( p(n) \). Thus, we have:
   \[
   \sum_{k=1}^{n} q(n, k) = p(n)
   \]
3. To find the sum of the number of groups over all splittings, we need to consider the contribution of each \( k \) groups. For each splitting into \( k \) groups, there are \( k \) groups. Therefore, the sum of the number of groups over all splittings is:
   \[
   \sum_{k=1}^{n} k \cdot q(n, k)
   \]
4. Consider the function \( p(n+1) \), which represents the number of ways to split \( n+1 \) pieces of candy. Each splitting of \( n+1 \) pieces can be derived from a splitting of \( n \) pieces by either adding the \( (n+1) \)-th piece to one of the existing groups or by creating a new group with just the \( (n+1) \)-th piece.
5. Therefore, the number of ways to split \( n+1 \) pieces into \( k+1 \) groups is \( q(n, k) \), and the number of ways to split \( n+1 \) pieces into \( k \) groups is \( k \cdot q(n, k) \) (since the \( (n+1) \)-th piece can be added to any of the \( k \) groups).
6. Summing these contributions, we get:
   \[
   p(n+1) = \sum_{k=1}^{n} (k+1) q(n, k)
   \]
7. We can rewrite this as:
   \[
   p(n+1) = \sum_{k=1}^{n} k q(n, k) + \sum_{k=1}^{n} q(n, k)
   \]
8. Using the fact that \( \sum_{k=1}^{n} q(n, k) = p(n) \), we have:
   \[
   p(n+1) = \sum_{k=1}^{n} k q(n, k) + p(n)
   \]
9. Rearranging, we find:
   \[
   \sum_{k=1}^{n} k q(n, k) = p(n+1) - p(n)
   \]
10. For \( n = 8 \), we need to find the sum of the number of groups over all splittings of 8 pieces of candy:
    \[
    \sum_{k=1}^{8} k q(8, k) = p(9) - p(8)
    \]
11. From the given table, \( p(9) = 21147 \) and \( p(8) = 4140 \). Therefore:
    \[
    \sum_{k=1}^{8} k q(8, k) = 21147 - 4140 = 17007
    \]

The final answer is \(\boxed{17007}\).