Instead of confusing ourselves with the positions of the L's, we
only need to look at where the G and I will go, and the L's will
just go in the remaining positions. Starting with the G, we can put
it in one of 4 places. Then there are 3 remaining places to put the
I. So the number of words is $4 \times 3=12$.
This problem is taken from the UKMT Mathematical Challenges.