Where am I overcounting? 
How many 3-element subsets of $\{1,2,3,...,19,20 \}$ are there such that the product of the three numbers in the subset is divisible by $4$ 

My approach
Define the subsets $A_i = \{x\:\: \mid \: \: x \equiv i \pmod 4 \}$ for $0 \leq i \leq 3$
Case 1.
We have one from $A_0$ and two from anywhere else: $\binom{5}{1} \cdot \binom{19}{2}$
Case 2.
We have two from $A_2$ and one from anywhere except $A_0$: $\binom {5}{2} \cdot \binom{13}{1}$ 
We add these to get $985$. But the answer is $795$. Where am I overcounting? 
 A: It is easier to write $E=\{1,\ldots,20\}$ as $A\cup B\cup C$ where $A$ is the set of odd numbers in $E$, $B$ is the set of numbers of the form $4k+2$ in $E$ and $C$ is the set of numbers of the form $4k$ in $E$. We have $|A|=10, |B|=|C|=5$. Now we separate cases. For a subset $F$ of $3$ elements of $E$, the product of the elements of $F$ is a multiple of four iff
$|F\cap C|\geq 1$ or $|F\cap C|=0$ and $|F\cap B|\geq 2$. We have
$$ \binom{5}{1}\cdot\binom{15}{2}+\binom{5}{2}\cdot\binom{15}{1}+\binom{5}{3} =685$$
subsets of such kind such that $|F\cap C|\geq 1$, and
$$ \binom{5}{2}\cdot 10+\binom{5}{3}=110 $$
subsets of such kind such that $|F\cap C|=0$ and $|F\cap B|\geq 2$, hence the final answer is given by $685+110=\color{red}{795}$.
A: To avoid overcounting, we consider in 3 disjoint cases:
i) At least one in $A_0$, that is the complement of three in
$A_1\cup A_2\cup A_3$: $\binom{20}{3}-\binom{15}{3}=685$.
ii) Two in $A_2$ and one in $A_1\cup A_3$: $\binom{15}{2}\cdot 10=100$.
iii) Three in $A_2$: $\binom{5}{3}=10$.
Hence the total is $685+100+10=795$.
A: For yet another approach, consider tackling the problem from the opposite perspective: which combinations don't give multiples of $4$?  It's easy to see that any such must consist of either 'three odd numbers' or 'two odd numbers and one number $\equiv 2\pmod 4$', and that these sets are disjoint.  This gives ${10\choose 3}+{10\choose 2}\cdot{5\choose 1}$ $= 120+\cdot 45\cdot 5$ $=345$ subsets that don't multiples of $4$, and subtracting that from the ${20\choose 3}=1140$ total subsets yields the result of $795$.
A: The multiplicities of the prime factor $2$ are 
$$0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2$$
and the frequencies follow a geometric progression of ratio $1/2$, with rounding up or down:
$$\color{green}{10}\times0,\ \color{blue}{5}\times1,\ 3\times2,\ 1\times3,\ 1\times4.$$
Among the $\binom{20}3$ possible triples, there are $\binom{\color{green}{10}}3$ of them with multiplicity $0+0+0$, and $\binom{\color{green}{10}}2\binom{\color{blue}5}1$ with multiplicity $0+0+1$. All others have multiplicities $\ge2$.
Hence
$$1140-120-45\cdot5=795.$$
