Let $\mathbb F:=\mathbb F_5$ the field with five elements.
(i) How many elements has $\mathbb F^3$?
(ii) How many different basis has $\mathbb F^3$?
My idea:
(i) $\mathbb F^3$ has $5^3$ elements.
(ii) The 1. basis vector must be different from the zero vector, so i've got $5^3-1$ possible choices for $v_1$.The 2. basis vector $v_2$ may not be linear combination $v_2=a\cdot v_1$ with $a \in \mathbb F$, so $5^3-5$ possible choices, and for $v_3$ i've got $5^3-5^2$ possible choices, so altogether $(5^3-1)(5^3-5)(5^3-5^2)$. But i'm unsure, whether i've counted to many? Do i have to divide my result through $n!$?