combinatorics - arranging consecutive numbers One is arranging the $10$ numbers $0$ to $9$ in a row.
what is the probability of not having a consecutive of $7$ numbers or more?
for example, $2034567891$ is forbidden as it has $7$ consecutive numbers.
Edit :
What I have tried:
for $0-6,$ I treat it as one block along with $7,8,9.$ So the number of permutation for $0-6,7,8,9$ is $ 4!=24.$
For $1-7,$ we already counted $0-7$ for example so we need to consider it, so the number of uncounted permutations of $0,1-7,8,9$ is: $4!-3!=18 .$
Same for $2-8,$ (now we need to remove $1$ or $9$ or both). Number of permutations: $ 4!-3!-3!-2=12.$
For $3-9$ : $ 4!-3!=18.$
In total: 
$$ P = \frac{10!-72}{10!} = \frac{50399}{50400} $$
 A: I agree with most of the reasoning written out. For example, there's certainly $4!$ combinations involving $012...6.$ Likewise for those containing $1...7$ and not $0...6.$ 
The trouble comes with counting the permutations with $2...8.$ Here we want to count everything except what we've counted already, ie those with $0...6$ or $1...7.$ We don't want to count, say, $1234567809$ twice. So really we only need to avoid counting permutations with a $1$ before $2...8.$ So $4!-3!$ again. The argument is identical for $3...9,$ as long as $2$ isn't before $3...9$ it's a new sequence. 
This gives $24 + 3(18) = 78$ unwanted combinations. Giving a result of
$$\frac{10!-78}{10!}$$
A: Might not be the adaptable way to solve and certainly wouldn't be practical for just about any numbers but certainly the easiest is:
There is $1$ way to have all ten consecutive.
There are$2$ ways to have exactly nine ways consecutive: ($0$ to $8$ after a $9$ or $1$ to $9$ before a $0$.
exactly eight.  There are $3$ choices for the numbers: $0$ to $7$, $1$ to $8$, or $2$ to $9$.  $0-7$ can precede $9,8$; follow $9,8$ or $8,9$; or be between $8\*\*9$.  similarly $2-9$ can follow $1,0$; precede $0,1$ or $1,0$; or be between $0\*\*1$.  And $1-8$ can precede $0,9$; be between $9**0$; or following $0,9$.  That is $11$.
exactly $7$:  There $4$ choices for the numbers.  $0-6$ can be in $0,1,2,3,4,5,6,\*,\*,\*$ provided the next number isn't $7$; so there are $4$ ways to arrange $7,8,9$ so that it doesn't start with $7$.  Likewise for $1,2,3,4,5,6,7,\*,\*,\*$ and $2,3,4,5,6,8,\*,\*,\*$ provided the next numbers aren't $8$ or $9$. 
And the same is true for $\*,\*,\*,1,2,...,7$,  or $\*,\*,\*,2,....,8$, or $\*,\*,\*,3,....9$.
So that's $24$ more.
$3,4,5,6,7,8,9,\*\*\*$ and $\*,\*,\*,0,1,2,3,4,5,6$ can have $6$ ways each so that is $12$ more.
Now for $\*,a,b,c,d,e,f,g,\*,\*$ and $\*,\*,a,b,c,d,e,f,g,\*$.  So long as the it isn't preceded by $a-1$ and followed by $g+1$, this is good.  So for $--,0,.....,6,\*--$ or $-\*,3,....,9\*--$ there are $4$ ways (not followed by $7$ or not preceded by $2$ so that is $4*4 = 16$ more ways.
If $a\ne 0$ and $g \ne 9$ (i.e. $a= 1,2$ and $7,8$) then you you can arrange the three remaining numbers, $a-1, g+1, x$ as 1)$g+1$ before $a$ the $a-1$ after the $g$; 2)$g+1$ before the $a$ and $x$ after the $g$ 3)$x$ before the $a$ and $a-1$ after the $g$.  So that is $2\cdot 2\cdot 3=12$ more ways.
So there are $1+2 +11+(24 +12+16+ 12) = 78$ ways to have $7$ or more consecutive.
So there are $10! -78$ ways not to.
So the probability of not is $\frac {10!-78}{10!}$
THis really is not a good way to do it and it quickly becomes a nightmare unless you reign it in quickly with some recursion formulas.  But they look farirly simply.  Can't have $\*\*\*\*\*, a,b,....,h,\*\*\*\*$ preceded by $a-1$ and $h$ followed by $h+1$ and so.....
