3 digit numbers with conditions How many 3 digit natural numbers with distinct digits are there that have no consecutive digits (ascending or descending) ?
I have solved it by enumerating  the exclusions, but is there some slicker way ?
edit:
As suggested by one contributor, I am giving the answer, which is 399. That in any case can easily be checked by computer. The question I am posing is whether some slick combinatorial method exists for obtaining the answer. I just used commonsense categorisation of the exclusions, counted the # in each category and subtracted it from the number of distinct 3-digit #s.

It sounds like you are not allowing numbers like $182$ because the $1$ and $2$ are neighboring.  Let us first allow numbers to start with $0$.  Then we have to find three digits without adjacent ones, then multiply by $6$ for the orders.  Let there be $F(n)$ combinations of three digits out of $n$ possibilities without neighbors.  Also let $G(n)$ be the number of combinations of two digits without neighbors and $H(n)$ be combinations of one digit without neighbors.  Clearly $H(n)=n$.  For $G(n),$ the first element is either included or not.  If it is, we have $H(n-2)$ choices.  If not, we have $G(n-1)$ choices.  So $G(n)=n-2+G(n-1), G(2)=0.$   We find $G(n)=\frac 12(n-2)(n-1)$.  Then $F(n)=G(n-2)+F(n-1)=\frac 12(n-4)(n-3)+F(n-1), F(4)=0$ by the same reasoning.  This gives $F(n)=\frac 16(n-4)(n-3)(n-2)$
Now we can apply the leading zero restriction.  There are $F(9)=35$ combinations without zero, giving $210$ numbers and $G(8)=21$ choices with zero, giving $63$  choices.  The total I get is $273$.  Since this is not $399$, I am probably not counting the same thing you are.
Added:  based on Peter Phipp's comment, ignore the problem if the first and last digits are adjacent.  I started doing it by cases, but it wasn't slick.  The interaction of distinct digits across the number and no two adjacent made a mess.
