Ace high hands
$\displaystyle \dbinom{4}{1}*\dbinom{12}{4}*\left[\dbinom{4}{1}\right]^4=506880$
Is this correct because someone else said the answer is 502,860.
Hello, dwsmith!
Your answer is off . . . and Someone Else's answer is off, too.
Ace high hands
$\displaystyle \displaystyle \binom{4}{1}\cdot\binom{12}{4}\cdot\left[\binom{4}{1}\right]^4\:=\:506,\!880$
Is this correct because someone else said the answer is 502,860.
Your answer is correct . . . up to a point.
You said there are 4 choices for the Ace
. . then we select 4 of the other 12 values
. . and there are $\displaystyle 4^4$ 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: .$\displaystyle \{10,J,Q,K,A\}$
. . There are: .$\displaystyle {4\choose1}^5 \:=\:1024$ of them.
There are Ace-low straight: .$\displaystyle \{A,2,3,4,5\}$
. . There are: .$\displaystyle {4\choose1}^5 \:=\:1024$ of them.
Hence, there are: .$\displaystyle 1024 + 1024 \:=\:2048$ Straights.
For the Flushes, there are: $\displaystyle {4\choose1}$ choices for the Ace.
And $\displaystyle {12\choose4}$ choices for the other 4 cards of that suit.
Hence, there are: .$\displaystyle {4\choose1}\cdot{12\choose4} \:=\:1980$ Flushes.
It seems that there are: .$\displaystyle 1024 + 1980 \:=\:4028$ 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: .$\displaystyle 1980 + 2048 - 4 \:=\:4024$
Therefore, there are: .$\displaystyle 506,\!880 - 4,\!024 \;=\;502,\!856$ Ace-high hands.