How many ordered bases can I find? 
I have found a basis for this question, however I am curious how many correct solutions I could find? Can you explain how I would calculate this?
 A: You could find
$$3(p^3-p)(p^3-p^2)$$
ordered bases containing $(1,1,1)$. This is because there are 3 possible positions for $(1,1,1)$, then we need to choose a vector that is not in the subspace spanned by $(1,1,1)$, of which there are $p^3-p$, then we need to choose a vector not in the two dimensional subspace spanned by the vectors already chosen, of which there are $p^3-p^2$ since the subspace has $p^2$ elements.
A: Your vector space $V$ has $p^3$ elements. There are $p^3-p$ elements that are not scalar multiples of $(1, 1, 1)$, so there are that many ways to choose the second basis vector. Then, the two already chosen vectors span a subspace with $p^2$ elements, so there are $p^3-p^2$ ways to pick the last element. Therefore, there are $(p^3-p)(p^3-p^2)$ ordered bases with $(1, 1, 1)$ as the first vector. Since we're also allowed to have $(1, 1, 1)$ be the second or third vector in the basis, we multiply this by $3$ to get the total number of ordered bases containing $(1, 1, 1)$ as $3(p^3-p)(p^3-p^2)$. 
