$|V|=p^n$. An ordered <basis> is chosen successively in $(p^n-1)(p^n-p)\cdots(p^n-p^{n-1})$ ways, which also counts bijective endomorphisms. Counting ordered independent $k$-tuples and dividing by the number of <bases> of $\mathbb F_p^k$ gives the Gaussian binomial $\prod_{i=0}^{k-1}(p^n-p^i)/(p^k-p^i)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 4 / 1E / Solution

 Solution (source code)