Probability of deck of cards such that each person receives one ace 
Suppose that a deck of 52 cards containing four aces is shuffled thoroughly and the cards are then distributed among four players so that each player receives 13 cards. Determine the probability that each player will receive one ace.

The answer to this is given as$$\frac{13^4}{\binom {52}4}$$
My doubt is the following:


*

*The book justifies ${\binom {52}{4}}$ as number of possible different combinations of the four positions in the deck occupied by 4 aces. That sounds like a case of arrangements to me, so shouldn't we think about permutations and not combinations if we are concerned about how the aces are to be arranged in the deck ?.

*Shouldn't the denominator be ${\binom {52}{13}}$ since you are choosing 13 cards for 4 people. 


There are 
$$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$$
ways to distribute $13$ cards to each of four people.
There are $4!$ ways to distribute the aces so that each person receives one and 
$$\binom{48}{12}\binom{36}{12}\binom{24}{12}\binom{12}{12}$$
ways to distribute the remaining cards so that each person receives twelve of them.  Hence, the desired probability is 
\begin{align*}
\frac{4!\dbinom{48}{12}\dbinom{36}{12}\dbinom{24}{12}\dbinom{12}{12}}{\dbinom{52}{13}\dbinom{39}{13}\dbinom{26}{13}\dbinom{13}{13}} & = \frac{4! \cdot \dfrac{48!}{12!36!} \cdot \dfrac{36!}{12!24!} \cdot \dfrac{24!}{12!12!} \cdot \dfrac{12!}{12!0!}}{\dfrac{52!}{13!39!} \cdot \dfrac{39!}{13!26!} \cdot \dfrac{26!}{13!13!} \cdot \dfrac{13!}{13!0!}}\\[2mm]
& = \frac{4! \cdot \dfrac{48!}{12!12!12!12!}}{\dfrac{52!}{13!13!13!13!}}\\[2mm]
& = \frac{4!48!}{12!12!12!12!} \cdot \frac{13!13!13!13!}{52!}\\[2mm]
& = \frac{4!48!13^4}{52!}\\[2mm]
& = \frac{13^4}{\dfrac{52!}{4!48!}}\\[2mm]
& = \frac{13^4}{\dbinom{52}{4}}
\end{align*} 
Let's compare this solution with the approach of your author.  As you stated, there are $\binom{52}{4}$ ways to choose the four positions occupied by the aces in the deck.  Since each person receives $13$ cards, there are $13$ possible places for the position of the ace in each person's hand.  Hence, the desired probability is 
$$\frac{13^4}{\dbinom{52}{4}}$$ 
