How many 13 hand cards have one ace? I know the answer is ${4 \choose 1}{48 \choose 12}/{52 \choose 13}$. But I have trouble rationalizing it. Why is it not ${13 \choose 1}{4 \choose 1}{48 \choose 12}/{52 \choose 13}$ as in choose which of the $13$ spots the ace goes into and then choosing which of the $4$ aces and then choosing the rest of the twelve hands? I know it's wrong, but why is it wrong to think like this? Sorry I am really bad at combinatorics. Thanks.
 A: Remember, when it comes to hands, for example a five card hand
$A2346$ is the same as $6A234$. In other words, all $5!$ arrangements are equivalent. So, it does not matter  which 'spot' the Ace goes in, for example. We just care that the Ace is in our hand.
Hence, for a 13 card hand, we have to choose one ace, and there are $\binom{4}{1}$ ways to do it. Next, we choose the rest of the cards. Since they cannot be aces, then we have $52-4 = 48$ cards to choose from. So there are $\binom{48}{12}$ ways to choose the rest. Finally, there are $\binom{52}{13}$ ways to make a 13 card hand.
Thus the probability of making a hand with just one ace is
$$\frac{\binom{4}{1}\binom{48}{12}}{\binom{52}{13}}  =\frac{9139}{20825} = 0.438847539$$ 
Note $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Addendum:
I'm not too familiar with games using 13 card hands, like Bridge (I think). But I am familiar with 5 card poker hands. 
Let's say I want to calculate the probability getting a full house.
This would be something like $AAAKK$. Here is an instance where you do need $\binom{13}{1}$. 
First I choose the rank that I want to be the triple. There are 13 ranks and so there are $\binom{13}{1}$ ways to choose the rank for the triple. Notice that I am not counting where the triple goes, but which rank will be constitute the triple. Let's say I choose Ace. There are four aces (diamonds,spades, clubs, hearts) but I only need 3. There are $\binom{4}{3}$ ways to choose the three suits of the aces. Next I choose the other rank and which two suits will make the double. There are $\binom{12}{1}$ ways to choose the other rank, and $\binom{4}{2}$ ways to choose the suits. Finally, there are $\binom{52}{5}$ ways to make a hand. Hence, the probability of making a full house is
$$\frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}}{\binom{52}{5}} = \frac{6}{4165} = 0.0014405762.$$
A: The answer you claim to know is the fraction, not the number.
To determine the number you have to consider the difference of combinations and permutations. In the second case, order is important, in the first case not.
Only in the case where order is important, you'd have to multiply by 13.
The formulas for the number of combinations of card sets, you seem to understand perfectly.
A: Your heading talks of number of hands, but
if you are interested in the probability, you can use permutations,(though involving unnecessary complexity for this problem), but what you did was to use permutations for one card and combinations for the rest. 
If you use permutations whole hog, Pr $=\dfrac{52\cdot^{48}P_{12}}{^{52}P_{13}}$ will give the same result.
Or even the oddball approach of allowing choice of slots only to the first occupant, but in both numerator or denominator, would do, Pr $=\frac{52\cdot\binom{48}{12}}{13\cdot\binom{52}{13}},$
but needless to say, simplest is best, which here is to just use combinations. !
Added
If, as per the header, you are interested in the number of hands, a hand just doesn't consider the order in which the cards are dealt.  
A: You can reason as follows: One of the hands that interest you is made up of an ace and 12 non-aces. The ace is selected as $1$ in $4$, the non-aces as $12$ in $52 - 4$. So there are $\binom{4}{1} \binom{52-4}{12}$ such hands. In all, there are $\binom{52}{13}$ hands.
In general, to approach such complex situations, it pays off to look for a way to describe what you are trying to count, hopefully as a sequence of independent decisions (pick the ace, pick the other cards). Counting the number of sequences is easy. And it avoids stupid mistakes. Just take care that there is just one description for a particular object, or that each object has e.g. 2 descriptions.
Check the MIT lecture notes by Lehman, Leighton, and Meyer "Mathematics for Computer Science" (there are very often new editions, look around). Relevant is part III "Counting", in particular the examples with poker hands (section 15.7 in the current edition).
