My first combinatorics post!

The total number of $3$ -element subsets is $20 \choose 3$ $=1140$ .

Now, we count the $3$ -element subsets say ${a,b,c}$ such that $4$ does not divide $abc$ .

This is possible if and only if either all the three numbers are odd or any two of them are odd and the other an even number not divisible by $4$ .

There are $10$ odd numbers in the set ${1,2,3,\ldots,19,20}$ and $5$ even numbers not divisible by $4$ .

Thus the number of $3$ -element subsets ${a,b,c}$ such that $4$ does not divide $abc$ is $10 \choose 3$ $+5$ $10 \choose 2$ $=345$ .

Thus the number of $3$ -element subsets such that the product of these elements is divisible by $4$ is $1140-345= \boxed{795}$ .