What is the probability of no pair's names being adjacent? Suppose there are $N$ (even number, positive) people. And each one person has to find one and only one partner to form a pair. There is also a roster within which everyone's name appearing in alphabetic order. 
For example, if $N=4$, fours guys with last name initials A, B, C, D, their names appear as 
\begin{array}{c}
\text{*** A***} \\
\text{*** B***} \\
\text{*** C***} \\
\text{*** D***} \\
\end{array}
in the roster. Out of total number of pair-up schemes (AB)(CD), (AC)(BD), (AD)(BC) only (AC)(BD) is the nonconsecutive case of pairing up. So the probability in the case $N=4$ is 1/3.
Numerically I tested $N = 4$ through 16 with a computer programme. Here is the results,
\begin{array}{c|l|l|l}
\hline
N & \text{total pair-ups} & \text{nonconsecutive cases} & \text{probability} \\ \hline
4 & 3 & 1 & 0.3333333333 \\ \hline
6 & 15 & 5 & 0.3333333333 \\ \hline
8 & 105 & 36 & 0.3428571429 \\ \hline
10 & 945 & 329 & 0.3481481481 \\ \hline 
12 & 10395 & 3655 & 0.3516113516 \\ \hline
14 & 135135 & 47844 & 0.3540459540 \\ \hline
16 & 2027025 & 721315 & 0.3576133166 \\ \hline
\end{array}
Given $N$, the second column can be computed by 
\begin{align*}
  \text{Total combinations ($N$)} = \frac{{N \choose 2}{N-2 \choose 2}\cdots{4 \choose 2}{2 \choose 2}}{\left( \frac{N}2 \right)!}.
\end{align*}
So my question is, is it possible to compute (with a formula given $N$) the total number of all cases of pairing-ups where no pair's names would appear consecutively in the roster, hence the third column in the table? Also as $N \to \infty$, is there a non trivial limit (e.g. trivial limit 1) for the probability of the fourth column?
 A: Neat problem! As you rightly compute, for even $N$ the number of ways to pair up $N$ objects without any restrictions is $\frac{{{N}\choose{2}}{{N-2}\choose{2}}\cdots{{2}\choose{2}}}{\left(\frac{N}{2}\right)!}$. Let's call this number $a_N$.
In order to determine the number of pairings which do not have any pairs in roster order, we appeal to inclusion/exclusion. Define $b_{n,k}$ to be the number of ways to pick $k$ pairwise disjoint pairs of consecutive numbers from  $\{1\ldots n\}$. Then the total number of pairing schemes for $n$ objects with no consecutive objects paired is $\displaystyle\sum_{k=0}^{N/2} (-1)^ka_{N-2k}b_{N,k}$.
It turns out that $b_{n,k} = {{n-k}\choose{k}}$. Given the simple form, there must be a simple combinatorial proof, but I'm not seeing it right now. [NOTE: Simpler combinatorial proof appears in the comments, thanks to a tip from @joriki.] Here's an induction proof on $k$. Clearly $b_{n,0}=1$ for all $n \ge 0$, and it is easy to see that $b_{n,1}= n-1 = {{n-1}\choose{1}}$ for all $n \ge 0$. So suppose $b_{n,k-1} = {{n-k+1}\choose{k-1}}$ for all $n \ge 0$, and let's count the number of ways to pick $k$ pairwise disjoint pairs of consecutive numbers from $\{1\ldots n\}$. Each such picking has a smallest pair: if that smallest pair is $\{1,2\}$, then there are $b_{n-2,k-1}$ ways to pick $k-1$ additional pairwise disjoint pairs from the remaining $n-2$ numbers. If the smallest pair is $\{2,3\}$, then there are $b_{n-3,k-1}$ ways to pick the $k-1$ additional pairs, etc. Hence $b_{n,k} = \displaystyle\sum_{j=0}^{n-2} b_{j,k-1}$, which equals $\displaystyle\sum_{j=0}^{n-2} {{j-k+1}\choose{k-1}}$. Using the identity for Sum of Binomial Coefficients over the Upper Index, we obtain the result.
So for even $N$, the number of pairings of $N$ objects which do not have any pairs in roster order is $\displaystyle\sum_{k=0}^{N/2} (-1)^k\frac{{{N-2k}\choose{2}}{{N-2k-2}\choose{2}}\cdots{{2}\choose{2}}}{\left(\frac{N-2k}{2}\right)!}{{N-k}\choose{k}}$.
And of course, after I went through this, I thought to look in the Online Encyclopedia of Integer Sequences. It's pretty much there: http://oeis.org/A000806.
As far as an asymptotic limit for the probability of picking such a sequence, the calculation is actually not too bad. For any $n$, we simply divide the above by $a_N$ to get $$P_N = \displaystyle\sum_{k=0}^{N/2} (-1)^k \frac{\left(\frac{N}{2}\right)\left(\frac{N-2}{2}\right)\cdots\left(\frac{N-2k+2}{2}\right){{N-k}\choose{k}}}{{{N}\choose{2}}{{N-2}\choose{2}}\cdots{{N-2k+2}\choose{2}}}$$
Each summand is the ratio of two degree $2k$ polynomials in $N$, with leading coefficient $\frac{(-1)^k}{k!}$, so we have:
$$\displaystyle\lim_{N\rightarrow\infty}P_N = \sum_{j=0}^\infty \frac{(-1)^k}{k!} = \frac{1}{e}$$.
ADDENDUM: The OP asked for more detail regarding the inclusion/exclusion argument. Define $Z_N$ to be the set of all pairings of the integers $1 \ldots N$ which have at least one consecutive pair. Then we can write $Z_N = \displaystyle\bigcup_{i=1}^{N-1} Z_N^{i}$, where $Z_N^{i}$ is the set of all pairings containing the pair $\{i,i+1\}$.
The principle of inclusion/exclusion says that (for distinct indices) $$|Z_N| = \displaystyle\sum_{i_1\in \{1 \ldots N-1\}} |Z_N^{i_1}| - \displaystyle\sum_{i_1,i_2\in \{1 \ldots N-1\}} |Z_N^{i_1} \cap Z_N^{i_2}| + \displaystyle\sum_{i_1,i_2,i_3\in \{1 \ldots N-1\}} |Z_N^{i_1} \cap Z_N^{i_2} \cap Z_N^{i_3}| \ldots$$
For the first summation, $|Z_N^{i_1}|$ is the number of pairings in which $\{i_1,i_1+1\}$ form one of the pairs, which is clearly $a_{N-2}$, since we can arbitrarily pair the remaining integers. So this sum is $(N-1) a_{N-2} = a_{N-2}b_{N,1}$.
For the second summation, the size of the intersection $Z_N^{i_1} \cap Z_N^{i_2}$ is either $a_{N-4}$, if all of $i_1$, $i_1+1$, $i_2$, and $i_2+1$ are distinct, or $0$ otherwise. Hence the second summation is $a_{N-4}$ times the number of ways to pick $2$ pairwise disjoint pairs of consecutive integers in $\{1 \ldots N\}$, which is exactly $b_{N,2}$.
Reasoning analogously for the higher order intersections, we find $$|Z_n| = \displaystyle \sum_{k=1}^{N/2} (-1)^{k-1} a_{N-2k}b_{N,k}$$
But notice that $Z_n$ is the complement of the set we are interested, namely the number of pairings with no consecutive pairs, hence our solution is $$a_N - \displaystyle \sum_{k=1}^{N/2} (-1)^{k-1} a_{N-2k}b_{N,k}$$
which can be written as
$$\displaystyle \sum_{k=0}^{N/2} (-1)^{k} a_{N-2k}b_{N,k}$$.
