How many $13$ card hands have an ace and a king of the same suit?

Question

I have the following question, and my current attempt to the solution is:

Choose an ace out of $4$ aces and then choose one king and then choose $11$ cards randomly out of remaining $44$ cards, which is:

$$\binom41 \binom{44}{11}$$

Answer 1

An alternative to inclusion-exclusion is to first count the complementary number of ways to avoid having an Ace and King of the same suit, organizing this count according to the number of Aces that are chosen, which ranges from $0$ to $4$. In general, if you choose $k$ Aces, then, in order to avoid any Ace-King matches, the other $13-k$ cards need to be chosen from among the $48-k$ cards that are neither Aces nor the chosen Aces' matching Kings. The total complementary count is thus

$${4\choose0}{48\choose13}+{4\choose1}{47\choose12}+{4\choose2}{46\choose11}+{4\choose3}{45\choose10}+{4\choose4}{44\choose9}$$

and so the count you want is

$${52\choose13}-\left({4\choose0}{48\choose13}+{4\choose1}{47\choose12}+{4\choose2}{46\choose11}+{4\choose3}{45\choose10}+{4\choose4}{44\choose9}\right)$$

Note, however, that in this case the inclusion-exclusion approach, as in drhab's answer, involves fewer terms (four terms to be added or subtracted, instead of six).

Answer 2

Let $S$ denote the $13$ card hands having ace and king of spades.

Let $D$ denote the $13$ card hands having ace and king of diamonds.

Let $H$ denote the $13$ card hands having ace and king of hearts.

Let $C$ denote the $13$ card hands having ace and king of clubs.

Then by inclusion/exclusion and symmetry we find:$$|S\cup D\cup H\cup C|=\binom41|S|-\binom42|S\cap D|+\binom43|S\cap D\cap H|-\binom44|S\cap D\cap H\cap C|$$

Can you take if from here?