Combinatorial doubt, why ${}^{n}C_{1}.{}^{(n-1)}C_{1}\neq {}^{n}C_2$ 
Two cards are drawn simultaneously (or successively without replacement)
  from a well shuffled pack of 52 cards.

$$
\#(\text{one king and one non-king})={}^{4}C_{1}.{}^{48}C_{1}=\frac{4!}{3!*1!}.\frac{48!}{47!*1!}=4*48\\
\#(\text{no king})={}^{48}C_{2}=\frac{48!}{46!*2!}=\frac{48*47}{2}
$$
Fine, but why can't we do instead
$$
\#(\text{no king})={}^{48}C_{1}.{}^{47}C_{1}=\frac{48!}{47!*1!}.\frac{47!}{46!*1!}=48*47\neq {}^{48}C_2
$$
 A: When you do as in your second calculation, what happens is this:
Suppose you first choose spade $2$ (with $\binom{48}{1}$) and then choose spade $3$ (with $\binom{47}{1}$). But this is as same as choosing spade $3$ first and then spade $2$. So, we just count each case twice, which results in finding $2$ times the correct result.
A: LHS: choose one object from the set of $n$ objects, the choose another one from the same set withput the first object. The key here isthe word 'same', because this implies that object 1 and object 2 can be chosen in two different ways (or orders, if you wish) as they belong to the same set. 
RHS: choose two objects from the same set accounting for the order, i.e. reduce the number of ways to choose these objects by 2. This returns the unique number of such choices.
A: When you do this :
$\#(\text{no king})={}^{48}C_{1}.{}^{47}C_{1}$
You essentially introduce an extra condition and your question becomes :
What are the number of ways to select a card that is not a King and without replacing it select another card which is not a King (again).
The way this differs with the original question is : when you actually have 48 cards and you have to select two cards and hence the order doesn't matter so you use ${}^{48}C_{2}$, so essentially you are choosing two cards at a time which counts as single choice whereas with the other question (mentioned in bold above) you necessarily impose a restriction of choosing a single card first from a pool of $48$ cards and then choose another card from the reduced pool (which is now $47$). So you are now choosing two cards one by one which counts as two separate choices, hence the double value in your answer.
Take this example. You have a set of two spade symbols:
$\spadesuit_{1} \spadesuit_{2}$
$1)$ If you want to choose two spades at once from the above set then you have:
$\spadesuit_{1} \spadesuit_{2}$ - (only choice) - ${}^{2}C_{2}=1$
$2)$  Now if you want to choose a(any) diamond first out of the set then you have:
$\spadesuit_{1}$ - $1^{st}$ Choice
$\spadesuit_{2}$ - $2^{nd}$ Choice 
So it becomes $1+1=2$ Choices, i.e. ${}^{2}C_{1}=2$.
$3)$ Now suppose that you have already chosen a spade (say $\spadesuit_{1}$) and now you want to choose another spade from the remaining set of spades, you have:
$\spadesuit_{2}$ - (only choice) - ${}^{1}C_{1}=1$
So now if you use AND between condition $2)$ and $3)$ you get the result as ${}^{2}C_{1}*{}^{1}C_{1}=2$ and clearly ${}^{2}C_{1}*{}^{1}C_{1} \ne {}^{2}C_{2}$
