The vector space $V(n,d)$ has dimension $n$ and the vector coordinates are from set $\Sigma$.
How many vectors can the vector space contain? Why?
If $\rm V$ is an $\rm n$-dimensional vector space over a field $\rm F$ then it has cardinality $\rm |V| = |F|^n$ if both $\rm |F|$ and $\rm n$ are finite; otherwise $\rm |V| = n\:|F| = max(n,|F|)\:$, as follows from basic properties of cardinal arithmetic. For example, this implies that the dimension of $\mathbb R$ over $\mathbb Q\:$ is $\rm |\mathbb R| > |\mathbb Q| = |\mathbb N|\:$.
V(n,d) contains dn vectors. As you know, each vector in V(n,d) can be described as an n-tuple (v1, ..., vn) where each coefficient vj is drawn from the coefficient-set Σ. It is straightforward to show that the number of such tuples is |Σ|n = dn.
In particular: if Σ is a finite field (such as the integers modulo 2), then the vector space has finitely many elements. Otherwise, it has infinitely many elements.