Count the number of decks of cards, where no king is on top of the ace of the same suit.

Let the sets $S_1, S_2 S_3$ and $S_4$ be the decks where king $i$ is on top of ace $i$. Then 
\[|S_i|=51!      \]
\[|S_i \cap S_j|=50! ,   i\neq j,\]
\[|S_i\cap A_j \cap S_k|=49!,     i\neq j, j\neq k, i \neq k,\]
and
\[|S_1 \cap S_2 \cap S_3 \cap S_4|=48!\]
Hence
\[|S_1 \cap S_2 \cup S_3 \cup S_4|={4\choose 1} 51! - {4 \choose 2} 50!+{4 \choose 3}49! -{4 \choose 4}48!\]
\[=4 \cdot 51!-6 \cdot 50!+4\cdot 49!-48!\]
which is the number of decks where at least one king is on top of the ace of the same suit. Hence the number of decks of cards, where no king is on top of the ace of the same suit is 
\[52! - (4 \cdot 51!-6 \cdot 50!+4\cdot 49!-48!). \]