Math problem

The answer is exactly 200. (Variation on Pascal's Triangle)

I get 32.

The problem is typical of a "how many routes through a road network without going back on yourself" problem. Each letter is a road junction with a left/right decision. The solution is found by cummulatively summing the possible routes to each junction as you go. For example the first three G's have value 1,2,1 respectively, which then gives the four I's as 1,3,3,1. The next line of G's will therefore be 4,6,4 and so on. (Check out Pascal's Triangle).

Yes well that explains it, now I get 33.

L=1
OO=1,1
GGG=1,2,1
IIII=1,3,3,1
GGG=4,6,4
RR=10,10
III=10,20,10
LLLL=10,30,30,10
LLL=40,60,40
EE=100,100
S=200

Each number is the sum of the preceding left and right "ears" above it.

...for example the first GGG = 1,2,1 because there is only one path to the outer G's but two to the middle one. Similarly, the line below is 1,3,3,1 because the second and third I's can be reached by 1+2 =3 ways from the G's above them, whereas the outer I's can only be reached by the single path G above them.

Good work fresco and thanks for the derivation of the answer.