8 cards are drawn without replacement. Find the probability of obtaining exactly $3$ aces, or exactly $3$ kings, or both. 
$8$ cards are drawn without replacement from an ordinary deck. Find the probability of obtaining exactly $3$ aces, or exactly $3$ kings, or both.

The answer is $\frac{2~\binom 4 3 \binom {48} 5 - \binom 4 3 \binom 4 3\binom {44} 2  }{ \binom {52} 8 }$
I am trying hard to understand the reasoning behind this answer but still don't fully get it
Can someone explain please. Thank you very much.
please read $\binom m n$ as "from m pick n"

Things I believe I understand:
$\binom {52} 8\,\leftarrow$ obvious. # of possible different 8 card hands
$2~\binom 4 3 \binom {48} 5\,\leftarrow$ the act of taking 3 figures of the same rank (be it aces or kings) and multiplying it by the way in which the 5 remaining cards can be chosen.
Thanks again.

It's the principle of inclusion and exclusion at work.
$$\mathsf P(A\cup K) = \mathsf P(A)+\mathsf P(K)-\mathsf P(A\cap K)$$
$\mathsf P(A)~=~{\binom 4 3\binom {48} 5}\big/{\binom {52} 8}$ is the probability of obtaining 3 aces and 5 non-aces.   Some of those 5 non-aces may be kings.
$\mathsf P(K)~=~{\binom 4 3\binom {48} 5}\big/{\binom {52} 8}$ is the probability of obtaining 3 kings and 5 non-kings.   Some of those 5 non-kings may be aces.
$\mathsf P(A\cap K)~=~{\binom 4 3\binom 4 3\binom {44} 2}\big/{\binom {52} 8}$ is the probability of obtaining 3 aces, 3 kings, and 2 cards of other faces.   However we've counted this event among both the above.   So to avoid over counting we subtract one occurrence.
$$\mathsf P(A\cup K) = \dfrac{\binom 4 3\binom {48} 5 + \binom 4 3\binom {48} 5- \binom 4 3\binom 4 3\binom {44} 2}{\binom {52} 8}$$
That is all.
$\Box$
