Number of 5 high lowball hands in a 7 card hand I am looking at a probability breakdown of lowball hands (lowest $5$ distinct cards, $12345$ being the best) here: http://www.durangobill.com/LowballPoker/Lowball_Poker_7_cards.html
The website lists the total number of $5$ high hands as $781,824$. Using a standard deck, I thought that the counting methodology would be $(4C_1)^5 \cdot (47C_2)$. This overestimates the amount of hands however. Why is this not the correct way to count? 
 A: You're double (and triple, and quadruple) counting because sometimes one or both of the $2$ cards you choose out of $47$ are lower than a 6.
For example, the hand consisting of Ace through 5 of diamonds and the Ace and 2 of hearts is counted 4 times.
The hand consisting of Ace through 5 of spades plus both red Aces is counted 3 times.
The hand consisting of 2 black Aces and the 2 through 6 of hearts is counted twice.

To set this up right, let's partition the cards: 20 cards are less than 6, 32 cards are 6 or more.
First let's count all the hands we don't need to double count or half count:
$${4 \choose 1}^5{32\choose2}$$
Next let's tackle the hands with 2 of some value below 6 (the value to be doubled is chosen out of five possible values):
$${5\choose1}{4\choose2}{4\choose1}^4{32\choose1}$$
Now let's tackle the hands with 3 of some value below 6 (note the 32 choose 0 included for completeness):
$${5\choose1}{4\choose3}{4\choose1}^4{32\choose0}$$
Now let's tackle the hands with 2 distinct values below 6 for which there are 2 cards:
$${5\choose2}{4\choose2}^2{4\choose1}^3{32\choose0}$$
Adding these figures together, we get 781824, just as stated on the page you linked.
A: The count can be organized as follows . . .

If a rank is at most $5$, call it a low rank.

If a hand contains the ranks $1,2,3,4,5$, call it a low hand.

Let 

$\;\;{\small{\bullet}}\;\;x_0$ be the number of low hands with no low rank duplicated.

$\;\;{\small{\bullet}}\;\;x_1$ be the number of low hands with exactly one low rank duplicated.

$\;\;{\small{\bullet}}\;\;x_2$ be the number of low hands with two low ranks duplicated.

$\;\;{\small{\bullet}}\;\;x_3$ be the number of low hands with some low rank triplicated.

Then

$x_0 = {\large{{\binom{4}{1}}^5\binom{32}{2}}}=507904$.

$\qquad$Explanation:

$\qquad{\small{\bullet}}\;\;$Choose the $5$ low rank cards:
$\;{\binom{4}{1}}^5\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the $2$ remaining cards, not of low rank:
$\;{\binom{32}{2}}\;$choices.

$x_1 = {\large{\binom{5}{1}\binom{4}{2}{\binom{4}{1}}^4\binom{32}{1}}}=245760$.

$\qquad$Explanation:

$\qquad{\small{\bullet}}\;\;$Choose the duplicated low rank:
$\;\binom{5}{1}\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the $2$ cards for that rank:
$\;\binom{4}{2}\;$choices

$\qquad{\small{\bullet}}\;\;$Choose the other $4$ low rank cards:
$\;{\binom{4}{1}}^4\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the remaining card, not of low rank:
$\;\binom{32}{1}\;$choices.

$x_2 = {\large{\binom{5}{2}{\binom{4}{2}}^2{\binom{4}{1}}^3}}=23040$.

$\qquad$Explanation:

$\qquad{\small{\bullet}}\;\;$Choose the $2$ duplicated low ranks:
$\;\binom{5}{2}\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the $2$ cards for each of those ranks:
${\;\binom{4}{2}}^2\;$choices

$\qquad{\small{\bullet}}\;\;$Choose the other $3$ low rank cards:
$\;{\binom{4}{1}}^3\;$choices.

$x_3 = {\large{\binom{5}{1}\binom{4}{3}{\binom{4}{1}}^4}}=5120$.

$\qquad$Explanation:

$\qquad{\small{\bullet}}\;\;$Choose the triplicated low rank:
$\;\binom{5}{1}\;$choices.

$\qquad{\small{\bullet}}\;\;$Choose the $3$ cards for that rank:
$\;\binom{4}{3}\;$choices

$\qquad{\small{\bullet}}\;\;$Choose the other $4$ low rank cards:
$\;{\binom{4}{1}}^4\;$choices.

Then the total number of low hands is
$$x_0 + x_1 + x_2 + x_3 = 507904 + 245760 + 23040 + 5120 = 781824$$
In your count of$\;{\large{{\binom{4}{1}}^5\binom{47}{2}}}=1106944$,

$\qquad{\small{\bullet}}\;\;$Each $x_0$-type hand was counted correctly.

$\qquad{\small{\bullet}}\;\;$Each $x_1$-type hand was counted $2$ times.

$\qquad{\small{\bullet}}\;\;$Each $x_2$-type hand was counted $4$ times.

$\qquad{\small{\bullet}}\;\;$Each $x_3$-type hand was counted $3$ times.

As a check:

$$x_0 + 2x_1 + 4x_2+3x_3 =  507904 + (2)(245760) + (4)(23040) + (3)(5120) = 1106944$$
which was the count you obtained.
