Q:

What's the probability of drawing all ranks in a standard 52-card deck?

Suppose a player is dealt with 16 cards under a standard 52-card deck, it's easy to find all possible combinations, which is ${52}\choose{16}$. 
However, I am struggling on finding all possible combinations that those 16 cards contains all ranks. 
And also wondering is there any general formula for finding such combination when cards dealt is more than 13?

A:

This answer works under the convention that for nonnegative integer $m$ and integer $i$ we have: $$i\notin\left\{ 0,\dots,m\right\} \implies\binom{m}{i}=0$$

Number the ranks with $1,2,\dots,13$ and let $R:=\left\{ 1,2,\dots,13\right\} $.
Define the random set $\hat{R}\subseteq R$ by stating that $r\in\hat{R}$
if and only if a card with rank $r$ is dealt.
Then to be found is: $$P\left(\hat{R}=R\right)$$

By means of inclusion/exclusion we find the (nice) general expression:$$P\left(\hat{R}=R\right)=P\left(\hat{R}\subseteq R\right)-P\left(\bigcup_{r\in R}\left\{ \hat{R}\subseteq R-\left\{ r\right\} \right\} \right)=$$$$\sum_{T\subseteq R}P\left(\hat{R}\subseteq T\right)\left(-1\right)^{\left|R\right|-\left|T\right|}\tag1$$
In the knowledge that $R$ has cardinality $13$ we find the (a bit less general) expression:$$P\left(\hat{R}=R\right)=\sum_{T\subseteq R}P\left(\hat{R}\subseteq T\right)\left(-1\right)^{13-\left|T\right|}\tag2$$
If $n$ cards are dealt (so in your question $n=16$) then: $$P\left(\hat{R}\subseteq T\right)=\binom{52}{n}^{-1}\binom{4\left|T\right|}{n}$$
Observe that $P\left(\hat{R}\subseteq T\right)$ only depends
on the cardinality of $T$ so that finally we can refine $(2)$ by: $$P\left(\hat{R}=R\right)=\binom{52}{n}^{-1}\sum_{k=0}^{13}\binom{13}{k}\binom{4k}{n}\left(-1\right)^{13-k}\tag3$$

