<s>[INST] Probability when arranging $N$ different numbers in a line we would like to arrange the numbers: $ 1, 2, \ldots , N$ in a line. 


*

*what is the probability that digits $1, 2$ are not arranged next to each other?

*what is the probability that digits $1, 2$ are not  arranged next to each other, and also the digits $3, 4$ are arranged next to each other?
my attempt:
This is mostly a combinatorics question, as the "Probability", meaning: $$\Omega = N!$$


*

*I treated $1, 2$ as a unit, looking for the number of different ways to arrange them next to each other - which would be the Complementary event.
$$P(A)^C = \frac{(N-1)! \cdot 2 }{N!}$$
where there's in fact $N-1$ needed to be arranged, ans 2 is the number of permutations.


so that $$P(A) = 1- \frac{(N-1)! \cdot 2 }{N!} = 1 - \frac{2}{N}$$
for the second sections - I'm not sure.
suppose I'd be looking the number of different ways to arrange  $1, 2, \ldots N$ in a line so that the digits $1, 2$ and $ 3, 4$ are always arranged next to each other, then that will be:
$$\frac{(N-2)! \cdot 2 \cdot 2}{N!}$$ where 4 is the number of permutations (2 for each unit of digits)
but does that mean that there's $$\left(N! - {(N-2)! \cdot 2 \cdot 2}\right) \cdot \left((N-2)! \cdot 2 \right)$$ ways to arrange the numbers? 
that doesn't seem right. 
 [/INST] Your answer to the first question is correct.

If the integers $1, 2, 3, \ldots, N$ are arranged in a line, what is the probability that the numbers $3$ and $4$ are arranged next to each other but the numbers $1$ and $2$ are not arranged next to each other.

As you observed, there are $N!$ ways to arrange the numbers in a row.
For the favorable cases, we begin by placing the numbers $3$ and $4$ in a box and arranging the $N - 3$ objects consisting of the box and the $N - 4$ numbers larger than $4$ in a row.  The objects can be arranged in $(N - 3)!$ ways.  Doing so creates $N - 2$ spaces in which we can place the $1$ and $2$, $N - 4$ between successive objects and two at the ends of the row.
$$\square \boxed{3, 4} \square 5 \square 6 \square \ldots \square N - 1 \square N \square$$ 
To separate $1$ and $2$, we must place them in two of these $N - 2$ spaces.  There are $N - 2$ ways to place $1$, which leaves $N - 3$ ways to place $2$.  The numbers $3$ and $4$ can be arranged within the box in $2!$ ways.  Hence, the number of arrangements of the $N$ positive integers $1, 2, 3, \ldots, N$ in which $3$ and $4$ are next to each other but $1$ and $2$ are not next to each other is 
$$(N - 3)!2!(N - 2)(N - 3) = (N - 2)!2!(N - 3)$$
Therefore, the probability that $3$ and $4$ are next to each other but $1$ and $2$ are not next to each other is 
$$\frac{(N - 2)!2!(N - 3)}{N!}$$
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