Probability that a 5-card hand contains exactly four cards of the same rank 
Given a traditional deck of $52$ cards, find the probability that a
  $5$-card hand contains four cards of the same rank.

So far, we only dealt with the notion of a finite sample space and that each elementary event is equally probable. So I defined the sample space 
$$ \Omega := \{\omega : \omega = (a_1,a_2,a_3,a_4,a_5), a_j \neq a_k, j\neq k, a_i \in \{1,\dots,52\}\}$$ since the sampling is without replacement. Now I am trying to figure out $$|\{\text{exactly four cards of the same rank}\}|$$ since then we have $$P(\{\text{exactly four cards of the same rank}\}) = \frac{|\{\text{exactly four cards of the same rank}\}|}{|\Omega|}$$ Does anyone has a nice way of calculating the cardinality of the above set? I tried to split it up into several disjoint subsets, but I do think this gets quite complicated.
 A: It's easy to see that there are 2 ranks (1 with 4 cards and 1 with 1).
There are $2!\frac{13!}{11!2!}=(13)(12)$ possibilities to choose the ranks. Then there are $4$ possibilities to choose the lone card's type (i.e. spade, heart, ...). So there are $(13)(12)(4)=(13)(48)$ possibilities. Multiply this by $5!=120$ to arrange (so far we have only chosen). We obtain $(13)(48)(120)=(52)(48)(10)(3)$. Divide this by the sample space $(52)(51)(50)(49)(48)$. Now we obtain $$\frac{1}{(17)(5)(49)}=\frac1{4165}\text.$$
A: There will be 13 different cases, one for every existing rank.
$${13 \choose 1} = 13$$
For every case you have that one of the cards can be whatever other card with a diferent rank, $52-4 = 48$, and the rest need to be one of the 4 cards of the same rank no matter the order in which they are picked. Therefore:
$${48 \choose 1} {4 \choose 4} = 48 $$
And in total that leads to $13\cdot48 = 624$ possible cases.
To calculate the probability you divide the favorable cases with the total possible states, i.e. all the possible 5 cards combinations taken randomly out of the maze:
$${52 \choose 5}  = \dfrac{52\cdot 51\cdot 50 \cdot 49 \cdot 48}{5!} = 2,598,960$$
$$\dfrac{624}{2,598,960} \approx 0.00024 $$
