Let $A$ be an infinite set of cardinality $\kappa$, and let nonzero $\lambda\leq\kappa$. Show that $|\{X\subseteq A~:~|X|=\lambda\}|=\kappa^\lambda$.
First, I tried to make an injection from $\{X\subseteq A~:~|X|=\lambda\}$ to $C = \{\text{set of functions from }B\text{ to }A\}$, $B$ is any set which has cardinality $\kappa$. Then I can show that $|\{X\subseteq A~:~|X|=\lambda\}|\leq\kappa^\lambda$.
Then I think I should consider the inclusion map from $C$ to $\{R \subseteq B \times A~:~|R| = \lambda\}$, and show that $|\{X\subseteq A~:~|X|=\lambda\}|\geq\kappa^\lambda$.
From above I drew direction, but I cannot think specifically. Can you help me to consider the injection in the First, and inclusion map in the second? thank you for your help.
A common notation for the set $\{X\subseteq A~:~|X|=\lambda\}$ is $[A]^\lambda$, which I will use in this answer. We are only interested in cardinality, so we could assume without loss of generality that $A=\kappa$ and $B=\lambda$.
The injection to show that $|[\kappa]^\lambda|\leq \kappa^\lambda$ needs to map each subset $X\subseteq\kappa$ of size $|X|=\lambda$ to a unique function $f_X:\lambda\to\kappa$. Since $|X|=\lambda$, there exists a bijection $f_X':\lambda\to X$, can you see how this bijection helps you to define $f_X$? Can you see why the map $[\kappa]^\lambda\to\kappa^\lambda$ that sends $X\mapsto f_X$ is an injection?
For the other direction, as you suggest, $f:\lambda\to\kappa$ is a relation on $\lambda\times\kappa$ (thus $f\subset\lambda\times \kappa$), and since $f$ has a domain of size $\lambda$ we have $|f|=\lambda$. Therefore, each $f:\lambda\to\kappa$ is an element of $[\lambda\times\kappa]^\lambda$, and thus we have an injection by considering the inclusion map $\kappa^\lambda\to[\lambda\times\kappa]^\lambda$. But now we're done, because $|[\lambda\times\kappa]^\lambda|=|[\kappa]^\lambda|$ (why?).