2 added explanation for why nCk is incorrect

This answer is similar to what Ben said about $d \times k$-matrices, but maybe a little more visual. Instead of computing the dimension "as a whole", I'll just compute the dimension of subspaces neighboring a given one.

Take some fixed $k$-dimensional subspace $P$ with complementary space $P^\perp$ of dimension $n-k$. Any sufficiently nearby subspace $P'$ to $P$ looks like a graph of a linear function $A : P \to P^\perp$. And on the other hand, any such linear function defines a unique subspace. So you only need to count linear maps from $P$ to $P^\perp$, which are $k(n-k)$-dimensional.

Thinking about the issue locally also helps avoid a mistaken dimension count like $n \choose k$. The problem with the $n \choose k$ count is that you are enumerating some random points on the Grassmannian, which doesn't tell you anything about the dimension. For example, take the case of 2-dimensional subspaces of $\mathbb{R}^3$. A subspace here is determined by its normal vector, so there is a bijection between 2-dimensional (oriented) subspaces and the unit sphere. The $3 \choose 2$ planes in your count correspond to the north pole and two equatorial points $90^\circ$ apart. But you wouldn't conclude that the sphere is 3-dimensional just because it has three points!

1

This answer is similar to what Ben said about $d \times k$-matrices, but maybe a little more visual. Instead of computing the dimension "as a whole", I'll just compute the dimension of subspaces neighboring a given one.

Take some fixed $k$-dimensional subspace $P$ with complementary space $P^\perp$ of dimension $n-k$. Any sufficiently nearby subspace $P'$ to $P$ looks like a graph of a linear function $A : P \to P^\perp$. And on the other hand, any such linear function defines a unique subspace. So you only need to count linear maps from $P$ to $P^\perp$, which are $k(n-k)$-dimensional.