The units digit of $$123456789\times8$$ is $$2$$, since $$9\times 8=72$$ . So, if the statement in the question is correct then the two digits which are in a different order are $$1$$ and $$2$$, whose sum is $$3$$. As a check, $$123456789\times8$$ is indeed $$987654312$$ . 
The result of the calculation $$123456789\times8$$ is almost the same as $$987654321$$ except that two of the digits are in a different order. What is the sum of these two digits? 

(a) $$3$$ 
(b) $$7$$ 
(c) $$9$$ 
(d) $$15$$ 