Trouble counting the number of "ace high" hands in poker I'm trying to count the number of "ace high" hands in a five card poker hand.  The solution from my answer key puts the count at 502,860; however, I have an argument for why this number is too high.  Please help me understand where my logic is flawed.
Instead of coming up with an exact answer for the number of ace high hands I will show an upper bound on the number of ace high hands.  First, go through the card deck and remove all four aces leaving a deck of 48 cards.  We will use this 48 card deck to form the four card "non ace" part of the "ace high" hand.  First, how many ways are there to form any four card hand from a 48 card deck?  This is (48 choose 4) = 194,580.  Now, not all of these hands when paired with an ace will form an "ace high" hand. For example A Q Q K K would be two pair.  In fact, any four card hand with at least two cards of the same rank (e.g. Queen of Spades, Queen of Hearts) will not generate an ace high hand.  So let's find the number of such hands and subtract them from 194,580.
I believe the number of such hands can be found by first selecting a rank for a pair from these 48 remaining cards, that is, (12 choose 1)--times the number of ways to select two suits for our rank (4 choose 2)--times the number of ways to pick the remaining 2 required cards from 46 remaining cards, that is, (46 choose 2).  So, restated, given our 48 card deck we can create a four card hand that contains at least one pair this many ways:
(12 choose 1)(4 choose 2)    (46 choose 2) = 74,520
   [pair rank]  [suits of pair] [remaining 2 cards]
Thus the number of four card hands that do not include at least one pair is:
(48 choose 4) - 74,520 = 120,060
We can pair each of these four card sets with one of our four aces to form the number of five card hands that contain an ace, but not any single pair (or better).  This is 120,060 * 4 = 480,240 hands.  However, this is already less than 502,860 shown by the key... and I haven't even begun to start subtracting out straights.  Clearly I have made a mistake, but what is it?

I came to 502,860;
Ace-High has no pairs, one card is an ace, and any other 4 cards
There are 12 other types of cards in the deck
So the total number of Ace high boards with any combination of the other 4 cards
is 12*11*10*9
there are 4*3*2 possible ways that the 4 non ace cards can be arranged
so there are a total of 12!/(8!4!) possible unique ace high hands
this comes to 495
2 of those possibilities contain straights AKQJT and A2345
so you subtract those 2; each card can be any of 4 suits, so there are 4^5 permutations of each unique ace high board
4 out of those 4^5 will be 5 cards of the same suit or a flush, so you subtract those 4. you then multiply this product by 493 and you get 502,860. formula is
[12!/(8!4!)-2]*(4^5-4)
