by Guest » Thu Aug 22, 2019 8:15 am
You are assuming that you can see both the face in front and the face behind? With n dice is a row, there are 2 dice on the two ends and n-2 "interior" dice. You can see 4 face of the two end dice (front, top, back and out side) but you can only see 3 faces of the "interior" dice (front, top, and back). So of n dice you can see 3(n- 2)+ 4(2)= 3n- 6+ 8= 3n+ 2 faces. Notice that with n= 3 that does give 3(3)+ 2= 9+ 2= 11. So with 2010 dice you would be able to see 3(2010)- 2.
(That looks like an old problem- lets update it: If you had 2019 dice how many faces could you see?)