# The number of subspaces of a vector space

Let $V$ be a vector space of dimension $n$ over $\mathbb{F}_q$, and let $U$ be a subspace of dimension $k$. I want to compute the number of subspaces $W$ of $V$ of dimension $m$ such that $W\cap U=0$.

I know why the number of subspaces of $V$ that contain $U$ and have dimension $m$ is $\binom{n-k}{m-k}_q$, but I don't understand why $q^{km}\binom{n-k}{m}_q$ is number of these subspaces?

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What does $[ \begin{array} nn-k \\m-k \end{array}]$ mean? –  Chris Eagle May 4 '12 at 15:43
@Chris I'd think that means the Gaussian binomial coefficient. So, $\binom n k _q$ will be the number of $k$ dimensional subspaces of $n$ dimensional vector space over the field of order $q$. –  user21436 May 4 '12 at 15:55
yes, this is Gaussian integer –  Babak Miraftab May 4 '12 at 16:25

Consider the standard projection $$\pi : V \to V/U,\quad v \mapsto v + U.$$ For a subspace $W$ of $V$, $W\cap U = \{0\}$ is equivalent to $\dim(\pi(W)) = \dim(W)$.
Using this characterization, you have to count the possible $\pi$-images of $W$ in $V/U$, and for a fixed such $\pi$-image $M$ the suitable $\pi$-preimages of $M$ in $V$.
Let me remark that in the spacial case $m = n - k$, you get the number of complements of $U$ in $V$ as $q^{k(n-k)}$.