Probability combinatorial question, random value switch I recently started studying probability from a textbook and in the textbook it gave a detailed breakdown of calculating the probability having a hand of cards (5) and X number of them being aces. however it seems to skip explaining a change in values.
The first step i completely understood:
P(AB)= P(A intersect B)
       P(B)

Which for this example are these values which make sense to me, 4 combinations of 2, and of the 5 card hand, 3 cards left can be any of the other 48 cards:
(4C2)(48C3)
  (52C5)

However in the next step in the textbook the values change to:
[(4·3)/(2·1)]·[(48·47·46)/(3·2·1)]
    (52·51·50·49·48)/(5·4·3·2·1)

Question
The step on the first line is the part i do not understand:
how has four combinations of 2 (4C2) become: 
(4 * 3)/(2*1)

Are they subtracting the combinations from the total number of possible cards to get the 3? and then why multiply the number 2 by 1?
I have looked around online for similar questions to see if they could elaborate on the steps in the textbook however they largely were using a number of different ways to calculate the probability and did not perform this step...
Thanks
 A: $n \choose k$ is the binomial coefficient. With it you can calculate the number of ways to choose a subset of k elements (2 aces) out of n elements (4 aces). The order is not regarded.
You have four aces (One for eauch suit). The suits are clubs (♣), diamonds (♦), hearts (♥) and spades (♠)
In how many was can 2 aces be chosen out of 4 aces, wihout regarding the order ? It can be written down. I only use the symbols for the different aces:
$\large{♣}$$\large{♦}$
$\large{♣♥}$
$\large{♣♠}$
$\large{♦♥}$
$\large{♦♠}$
$\large{♥♠}$
Two aces can be chosen out of 4 aces in 6 ways, if we disregard the order.
We can use the binomial coefficient as well.
${n \choose k}=\frac{n!}{k!\cdot (n-k)!}$, where $n!$ is the factorial of $n$
A nice interpretation of the formula you can be read here.
$(n-k)!$ can be cancelled out.
${n \choose k}=\frac{n\cdot (n-1)\cdot \ \ \ldots \ \ \cdot (n-k+1)}{k!}$
With $n=4$ and $k=2$ we have the binomial coefficient ${4 \choose 2}$
$n-k+1=4-2+1=3$ is the last factor in the numerator. Therefore
$\frac{n\cdot (n-1)\cdot \ \ \ldots \ \ \cdot (n-k+1)}{k!}={4 \choose 2}=\frac{4\cdot 3}{1\cdot 2}=6$
