Total number of $$6$$-digit number in which only and all the five digits $$1,3,5,7$$ and $$9$$ appear, is :
Correct option is C. $$\dfrac{5}{2}(6!)$$
$$1,3,5,7,9$$ $$\downarrow$$
selecting a digit which can be repeated $$=^5C_1$$
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Total number of $$6$$ digits $$=^5C_1\times \dfrac{6!}{2!}$$
$$=1800$$