Probability of two people getting all the spades in a group of four Four friends called Julia, Wayne, Tony and Joe are playing a card game. A standard pack of $52$ cards is dealt out, 13 cards being dealt to each player.  
What is the probability that the cards dealt to Julia and Wayne include all thirteen spades?  
The solution says this:
We split up the game into two hands, hand 1 with Julia and Wayne and hand 2 with Tony and Joe.
The total outcomes is $\binom{52}{26}\binom{26}{26}$
and the required ways is:
$$\binom{13}{13}\binom{39}{13}\binom{39}{13}$$  
However, I'm unsure how they got the desired ways to be that...
 A: For Julia and Wayne to have all of the spades, it must be the case that neither Tony nor Joe received any spades. Let us simply consider the chance of this latter scenario manifesting: We will deal out twenty six cards to Tony and Joe, and consider the chance that none of them is a spade:
The chance of the first card being a nonspade is $39/52$; for the second card, it is $38/51$; for the third card, it is $37/50$; and so forth. The resulting probability is:
$$\frac{39 \cdot 38 \cdots 14}{52\cdot51\cdots27} = \frac{26\cdot25\cdots14}{52\cdot51\cdots40} = \frac{19}{1160054}$$
A: That stated count should only have one factor of $\binom{39}{13}$. Perhaps it is a typo.
First of all, as it's written, $\binom{13}{13}\binom{39}{13}\binom{39}{13}$ is larger than the total outcome count of $\binom{52}{26}\binom{26}{26}$, so something is of course wrong.
To count "required ways", $\binom{13}{13}$ represents choosing which spades should go to Julia and Wayne (all 13 out of 13). The combined hand needs 13 additional cards from the remaining 39. So that explains a factor of $\binom{39}{13}$. If there is to be one more factor, it would be $\binom{26}{26}$, choosing which of the remaining unassigned 26 cards go to Tony and Joe (all 26).
