Question 1208704
<pre>
Case 1. 5 empty boxes.  So 1 of the 6 boxes must contain all 8 balls.
That's 6 ways.

Case 2. There are 4 empty boxes, and 8 balls total in the other 2 boxes.  

Subcase 2a: The other two contain 4 balls each. there are C(6,2)=15 
ways to choose 2 boxes to put them in.

Subcase 2b: The other two non-empty boxes contain 1&7, 2&6 or 3&5 balls.  
For each of those three distributions, choose a box for the larger number 
of balls 6 ways, then a box for the smaller number of balls 5 ways. 
That's 3*6*5=90 ways

That's 15+90=105 for case 2

Total number of cases: 6+105=111.

Edwin</pre>