Mind Boggling BALLS....

4+4^2+4^3+4^4+4^5+4^6+4^7+4^8=87380.

Split the problem into cases. The cases would be 1 ball in a row, 2 balls in a row, 3 balls in a row, and so on. There are 4 possible colors for each ball in the row. Therefore, there are 4 ways for the 1 ball in a row, 4x4 or 16 ways for the 2 balls in a row, 4x4x4 ways or 64 ways for the 3 balls in a row, and so on. Therefore, there are a total of 4+16+64+256+1024+4096+16384+65536=87380 ways.

I got the final result as following, the answer is $\sum_{a=1}^{8}\begin{pmatrix}\sum_{b=1}^{4}\begin{pmatrix} 4 \\ b \end{pmatrix}\begin{pmatrix}\frac{a!}{a_{1}!a_{2}!a_{3}!a_{4}!}\end{pmatrix}\end{pmatrix},$ where $b\leqslant a$ ,

$a_{1}+a_{2}+a_{3}+a_{4}=a$ ,

and $i\leqslant b$ , for $a_{i}$ .

Unfortunately, I cannot express it to be a more explicit form. I, however, obtained the answer as $\boxed{87380}$ which equal $\frac{4(1-4^8)}{1-4}$ .

PS. I'm not sure whether the answer should be 87381 since we are considering arrangements of at most 8 balls in a row, is zero-ball arrangement in?