5 cards are chosen from a standard deck. What is the probability that we get all four aces, plus the king of spades? We have $\binom{52}{5}$ ways of doing this. This will be our denominator.
We want to select all 4 aces, there are there are exactly $\binom{4}{4}$ ways of doing this.
Now we have selected 4 cards, and we need to select one more. That one card has to be the king of spades. Out of the 48 remaining cards, only one of them is our wanted one. This can be chosen in $\binom{48}{1}$ ways.
So is the answer:
$$\dfrac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}=\dfrac{1}{54145}?$$
Is this correct?
If we look at it another way:
Then then our probability for the aces and specific king are $(4/52)\times (3/51) \times (2/50) \times (1/49) \times (1/48)$ which is a completely different here.
Which is the correct approach?
 A: Your first computation gets the chance you get four aces and any other card.  The $48 \choose 1$ factor is selecting the other card.  If you want specifically the king of spades it should be $1$, so the chance is $48$ times less.  Your second computation gets the chance you draw the four aces first in some order, then draw the king of spades.  You should multiply by $5$ for the number of positions the king can be in.  Neither is correct for the question in the title.
A: The first way:
There is only one set of cards, that fits the specifications (all 4 aces and the king of spades) and $\binom{52}{5}$ different sets of cards that can all be drawn with equal probability. The chance to draw your set is
$$
\frac{1}{\binom{52}{5}}
$$
The second way:
The chance of drawing a specific card out of $n$ is $\frac{1}{n}$. So a specific sequence or order of 5 cards has the chance $\frac{1}{52}\cdot\frac{1}{51}\cdot\frac{1}{50}\cdot\frac{1}{49}\cdot\frac{1}{48}$ to be drawn. There are $5!$ different sequences or orderings of 5 cards, that countain the 5 specified cards (s. permutations). So the chance to draw the set is
$$
\frac{5!}{52\cdot51\cdot50\cdot49\cdot48} = \frac{5!\cdot 47!}{52!}
= \frac{1}{\frac{52!}{5!\cdot 47!}} = \frac{1}{\binom{52}{5}}
$$
For the last equality see Binomial coefficient.
A: You select 4 aces out of 4 possible aces and one king of spades out of 1 possible king of spades, so there is exactly 
$$\binom{4}{4}\binom{1}{1}=1$$
way of completing this set. Therefore this set is unique.
There are $\binom{52}{5}$ unique sets of five cards.
Therefore the probability of picking this set of five cards is:
$$P=\frac{\binom{4}{4}\binom{1}{1}}{\binom{52}{5}} = \frac{1}{2598960}$$
A: Assuming you just want those 5 cards in any order:
Since you have specified all 5 specific cards that you want, you don't need to consider the aces separately from the king. The probability of selecting these 5 cards is then
(Probability choosing any one of the 5) $\times$ (Probability of choosing one of the remaining 4) $\times$ ..., i.e. $$\frac{5}{52} \times \frac{4}{51} \times \frac{3}{50} \times \frac{2}{49} \times \frac{1}{48} = \frac{1}{54145} \times \frac{1}{48},$$ just to confirm what Ross Millikan said.
If, instead, you want the 4 aces before the king, we have $$\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49} \times \frac{1}{48}.$$
A: You've fully determined the five cards you want, so the probability is $\frac{1}{{52 \choose 5}}$ since there are ${52 \choose 5}$ different ways to choose 5 cards from the deck, and only one correct one.
