So, we have 6 distinguishable boxes A, B, C, D, E, F.


Case 1.  4 boxes of 6 are empty, two remaining boxes are not empty.

         There are 6*5 = 30 different ordered pairs to select 2 boxes from 6 boxes. 

           
                 +----------------------------------------+
                 |   We consider ordered pairs of boxes,  |
                 |   since the boxes are distinguishable. |
                 +----------------------------------------+


         So, there are 30 different possibilities to have an ordered pair of 2 concrete distinguishable boxes to fill.

         Let's calculate the number of ways we can distribute 8 undistinguishable balls in 2 selected boxes
         in a way that no one of these 2 boxes is empty.

         These ways are  (7,1), (6,2), (5,3), (4,4), (3,5), (2,6), (1,7).

         So, there are 7 different ways for each selected ordered pair of two boxes.

         Counting this way, we obtain, in total, 30 * 7 = 210 ways 
         to distribute 8 indistinguishable balls among 6 distinguishable boxes
         in a way that 4 boxes are empty; of the two remaining boxes no one is empty.


         Now, if to consider this way, for us the pair of boxes (A,B) with (7,1) balls 
         is the same as the pair (B,A) with (1,7) balls.

         Therefore, the number 210 should be divided by 2:  210 : 2 = 105.



Case 2.  5 boxes of 6 are empty; the remaining 6th box is filled with 8 balls.

         Obviously, there are 6 ways to select one box of 6 to fill it with 8 balls.


                There are no other cases to distribute.  
                Cases 1 and 2 exhaust all possible ways.


So, the  answer to the problem is the sum  105 + 6 = 111.


ANSWER.  There are 111 different ways to distribute 8 indistinguishable balls among 6 distinguishable boxes, 
         if at least four of the boxes must be empty.

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