Is there a simple formula for the cardinality of $\{A\subseteq\kappa\mid |A|\leq\lambda\}$ when $\lambda\leq\kappa$?

Question

If $\lambda\leq\kappa$ are infinite cardinals, how many subsets of $\kappa$ of size $\lambda$ are there? And of size $\leq\lambda$? Is there some sort of explicite formula for this? The internet isn't as helpful on this as one would expect (unless one knows what to look for, I guess).

Answer 1

We can easily calculate that.

Let $[\kappa]^\lambda$ be all the subsets of $\kappa$ of size exactly $\lambda$. Each one has a natural ordering, so there is an obvious injection into ${}^\lambda\kappa$, the set of all functions from $\lambda$ into $\kappa$, whose cardinality is exactly $\kappa^\lambda$.

On the other hand, every function from $\lambda$ to $\kappa$ is a subset of $\lambda\times\kappa$ of cardinality $\lambda$. Since the set $\lambda\times\kappa$ has cardinality $\kappa$, this gives an injection in the other direction. So we have: $$\left|[\kappa]^\lambda\right|=\kappa^\lambda$$

So all the subsets of size $\leq\lambda$ would be the set $$\bigcup_{\alpha<\lambda}[\kappa]^\alpha=[\kappa]^{\leq\lambda}.$$

Of course its cardinality is at most the union of $\lambda$ [disjoint] copies of $[\kappa]^\lambda$. But that would not increase the cardinality, so we again have $\kappa^\lambda$.

Finally, what is $\kappa^\lambda$ in more explicit terms? That depends on the universe of set theory we work in, and on $\kappa$ and $\lambda$ themselves. So there's no better answer without additional information.