My reasoning is:

I can have 1, 2 or 3 Reds.

Case 1 : 3 reds; there are 9 diff ways 3 reds can be on the fences, and only (2^2)-2 = 2 ways the rest can be Chosen, hence 9*2= 18

Case 2 : 2 Reds 10 diff ways , and (2^3)- 2 = 6 hence 10*6 = 60

Case 3 : 1 red Can be in 5 places, and (2^4)-2 = 14 hence 5*14 = 70