1. ## Counting

Ace high hands

$\dbinom{4}{1}*\dbinom{12}{4}*\left[\dbinom{4}{1}\right]^4=506880$

Is this correct because someone else said the answer is 502,860.

2. Hello, dwsmith!

Ace high hands

$\displaystyle \binom{4}{1}\cdot\binom{12}{4}\cdot\left[\binom{4}{1}\right]^4\:=\:506,\!880$

Is this correct because someone else said the answer is 502,860.

You said there are 4 choices for the Ace
. . then we select 4 of the other 12 values
. . and there are $4^4$ ways to select those values.

You found the number of hands which contain an Ace and no pairs. . Good!

However, this includes the hands that are Straights or Flushes.

There are Ace-high straights: . $\{10,J,Q,K,A\}$
. . There are: . ${4\choose1}^5 \:=\:1024$ of them.

There are Ace-low straight: . $\{A,2,3,4,5\}$
. . There are: . ${4\choose1}^5 \:=\:1024$ of them.

Hence, there are: . $1024 + 1024 \:=\:2048$ Straights.

For the Flushes, there are: ${4\choose1}$ choices for the Ace.
And ${12\choose4}$ choices for the other 4 cards of that suit.

Hence, there are: . ${4\choose1}\cdot{12\choose4} \:=\:1980$ Flushes.

It seems that there are: . $1024 + 1980 \:=\:4028$ Straights and Flushes.

And this is what Some Else subtracted.

But he/she forgot that we've overcounted.
There are some hands which are Straights and Flushes.

There are 4 hands that are Ace-high straight flushes.
(They're called Royal Flushes, remember?)

So, the number we subtract is: . $1980 + 2048 - 4 \:=\:4024$

Therefore, there are: . $506,\!880 - 4,\!024 \;=\;502,\!856$ Ace-high hands.