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?
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.
The other answers have explained why your solution overcounts, and derived the correct answer by ad hoc methods. Here is a solution using the in-and-out formula. The number of $7$-card hands containing a "wheel" (ace-to-five) is $$\binom50\binom{52}7-\binom51\binom{48}7+\binom52\binom{44}7-\binom53\binom{40}7+\binom54\binom{36}7-\binom55\binom{32}7=\boxed{781824}\ ;$$
$\binom50\binom{52}7$ is the total number of $7$-card hands;
$\binom51\binom{48}7$ is the number of ways you can choose a low rank and a $7$-card hand with no cards of that rank;
$\binom52\binom{44}7$ is the number of ways you can choose two low ranks and a $7$-card hand excluding both of those ranks, and so on.
P.S. Let $E$ be the set of all $7$-card hands. Let $A_i$ be the set of all $7$-card hands containing no card of rank $i;$ thus $A_1$ is the set of all $7$ card hands with no ace, etc. The set of all $7$-card hands containing a wheel is $$E\setminus(A_1\cup A_2\cup A_3\cup A_4\cup A_5).$$ By the in-and-out-formula, the number of hands containing a wheel is equal to $$|E\setminus(A_1\cup A_2\cup A_3\cup A_4\cup A_5)|$$ $$=|E|$$ $$-|A_1|-|A_2|-|A_3|-|A_4|-|A_5|$$ $$+|A_1\cap A_2|+|A_1\cap A_3|+|A_1\cap A_4|+|A_1\cap A_5|+|A_2\cap A_3|+|A_2\cap A_4|+|A_2\cap A_5|+|A_3\cap A_4|+|A_3\cap A_5|+|A_4\cap A_5|$$ $$-|A_1\cap A_2\cap A_3|-|A_1\cap A_2\cap A_4|-|A_1\cap A_2\cap A_5|-|A_1\cap A_3\cap A_4|-|A_1\cap A_3\cap A_5|-|A_1\cap A_4\cap A_5|-|A_2\cap A_3\cap A_4|-|A_2\cap A_3\cap A_5|-|A_2\cap A_4\cap A_5|-|A_3\cap A_4\cap A_5|$$ $$+|A_1\cap A_2\cap A_3\cap A_4|+|A_1\cap A_2\cap A_3\cap A_5|+|A_1\cap A_2\cap A_4\cap A_5|+|A_1\cap A_3\cap A_4\cap A_5|+|A_2\cap A_3\cap A_4\cap A_5|$$ $$-|A_1\cap A_2\cap A_3\cap A_4\cap A_5.$$ Since the deck has $52$ cards, and $4$ cards of each rank, we have: $$|E|=\binom{52}7;$$ $$|A_1|=|A_2|=|A_3|=|A_4|=|A_5|=\binom{48}7;$$ $$|A_1\cap A_2|=|A_1\cap A_3|=\cdots=\binom{44}7;$$ $$|A_1\cap A_2\cap A_3|=\cdots=\binom{40}7;$$ $$|A_1\cap A_2\cap A_3\cap A_4|=\cdots=\binom{36}7;$$ $$|A_1\cap A_2\cap A_3\cap A_4\cap A_5|=\binom{32}7;$$ and so $$|E\setminus(A_1\cup A_2\cup A_3\cup A_4\cup A_5)|=\binom{52}7-5\binom{48}7+10\binom{44}7-10\binom{40}7+5\binom{36}7-\binom{32}7$$ $$=\binom50\binom{52}7-\binom51\binom{48}7+\binom52\binom{44}7-\binom53\binom{40}7+\binom54\binom{36}7-\binom55\binom{32}7=\boxed{781824}$$
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.
