How many ways to put $10$ distinguishable balls in $5$ identical boxes so that no box has more than $3$ balls? As the title says, I was asked to count the number of ways one can put $10$ different balls inside $5$ identical boxes so that no box has more than $3$ balls. I am really confused by the restriction and I can't even compute the number without the restriction (It confuses me that the boxes can't be distinguished but the balls are). Any explanation is more than welcome.

A comparison with distinct balls in distinct boxes$\;$ vs$\;$ identical boxes yields a simple transformation (although $5$ cases can't be avoided here)
For the pattern $3-3-3-1-0,\;$ for distinct balls to distinct boxes, we have the product of two factorials,
[Fill Pattern]$\times$[permute Pattern]$\;=  
\large\frac{10!}{3!3!3!1!0!}\cdot\frac{5!}{3!1!1!}$
The simple transformation for identical boxes is to change the numerator of the second term to $1$,
thus $\quad\large\frac{10}{3!3!3!1!0!}\cdot\frac{1}{3!1!1!}$
So we simply apply this transform to each allowable pattern, viz.
$3-3-3-1-0:\;\;\large\frac{10!}{3!3!3!1!0!}\cdot\frac{1}{3!1!1!}$
$3-3-2-2-0:\;\frac{10!}{3!3~2!2!0!}\cdot\frac1{2!2!1!}$
$3-3-2-1-1:\;\frac{10!}{3!3!2!1!1!}\cdot\frac{1}{2!1!2!}$
$3-2-2-2-1:\;\frac{10!}{3!2!2!2!1!}\cdot\frac1{1!3!1!}$
$2-2-2-2-2:\;\frac{10}{2!2!2!2!2!}\cdot\frac1{5!}$
Computing and adding up, the answer is $\boxed{35245}$
