$$6$$ blue, $$7$$ green and $$10$$ white balls are arranged in a row such that every blue ball is between a green and a white ball. Moreover, a white ball and a green ball must not be next to each other. The number of such arrangements is

Each blue ball must be adjacent to one green and one white, while green and white balls cannot be adjacent directly. Counting the possible placements of the six blue positions and distributing the remaining green and white balls gives $$\binom62\binom93+\binom92\binom63=1980$$. Hence the number of arrangements is $$1980$$.