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.
 A: Since you are looking for numbers with 3 distinct digits, consider this card game that I have devised.
You have 10 cards labelled $0$ to $9$:
$$\boxed{1}\,\boxed{2}\,\boxed{3}\,\boxed{4}\,\boxed{5}\,\boxed{6}\,\boxed{7}\,\boxed{8}\,\boxed{9}\,\boxed{0}$$
...and 3 slots (hundreds, tens, ones) denoted by:
$$\heartsuit \,\spadesuit\, \clubsuit$$
There are $9$ ways to put a card in the $\heartsuit$ because you cannot use $\boxed 0$. Then you have $9$ cards remaining to put in the $\spadesuit$ and finally $8$ cards in the $\clubsuit$. This makes a total of $9 \times 9 \times 8=648$ numbers with 3 different digits.

To find the numbers with digits in ascending order, consider if you put $1$ in $\heartsuit$. You have $7$ cards to choose from $(2, 3, 4, ..., 8)$ to put at $\spadesuit$. 


*

*If you select $12\clubsuit$, you have $7$ choices $(3, 4, ..., 9)$ for $\clubsuit$

*If you select $13\clubsuit$, you have $6$ choices $(4, 5, 6, ..., 9)$ for $\clubsuit$


You can see the total number of numbers that begin with $1$ and have ascending digits is 
$$
\begin{align*}
7+6+5 +4+ \cdots + 1\\
+6+5+4+\cdots +1\\
+5+4+\cdots +1\\
\ddots \vdots\\
+1\\
=84
\end{align*}
$$
Do the same for digits that begin with $2$ at $\heartsuit$. You have $6$ cards to choose to put at $\spadesuit$ and subsequently, $6, 5, 4, ...$ cards to put at $\clubsuit$ depending on the number at $\spadesuit$.
The total number of numbers that begin with $2$ and have ascending digits is
$$
\begin{align*}
6+5+4+\cdots +1\\
+5+4+\cdots +1\\
\ddots \vdots\\
+1\\
=56
\end{align*}
$$
Do this until the last number, which is $789$. (You cannot put $8$ or $9$ in $\heartsuit$ or you would draw dead for the next 2 positions). There are a total of $7$ terms in this summation. The next numbers are $35, 20, 10, 6, 3, 1$ totalling $215$. 

For descending numbers, there $8$ ways to put a card in $\heartsuit$. You cannot put $0$; and putting $1$ will leave you drawing dead. If you put $9$ at $\heartsuit$, then you can choose from $8$ cards for the $\spadesuit$. You can see the pattern for the last digit...so let's get to the mathematics.
The total number of numbers that begin with $9$ and have descending digits is
$$
\begin{align*}
8+7+6+5 +4+ \cdots + 1= 36
\end{align*}
$$
Now, consider that you put $4$ at $\heartsuit$. You can pick from $3$ cards for the second position. Depending on the choice, you can pick $3, 2$ or $1$ card. The summation would be
$$
3+2+1 = 6
$$
The terms in this summation are $36, 28, 21, 15, 10, 6, 3, 1$ and this makes a total of $120$
I would put my money on $648-215-120 = 313$
A: 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.
A: I was hoping for a slicker method than what I had used, but as nothing has turned up, I am giving my method in the hope that someone may improve upon it.
Numbers can't start with 0, so total #s = 9*9*8 = 648
This part is easy.
It is easier to count exclusions rather than favorable cases
We need to exclude any 2 or 3 together
01 together: 2-9 preceding = 8 
10 together: 2-9 preceding or succeeding = 16 
12,23,........89 and 21,32, .......98 together (16 patterns)
remember, 0 can't precede, so
7 digits can precede, 8 succeed, => 16*15 = 240
But we are counting 123,234, ...789 & 321,432, .. 987, 
and also 210 twice, and we need to correct this
so total exclusions = 8+16+240 - 15 = 249
ans: 648 - 249 = 399 
