Given $n$ people, where $n$ is even, you can choose the first pair of people ${n \choose 2}$ ways, where ${n \choose 2}=\frac{n!}{2!(n-2)!}$. The next pair can be chosen in ${n-2\choose 2}$ ways, etc... The end result will be $\frac{n}{2}$ pairs which can be arranged in $\left(\frac{n}{2}\right)!$ ways. So there are $$\frac{{n \choose 2}{n-2 \choose 2}\dots{2 \choose 2}}{\left(\frac{n}{2}\right)!}=\frac{n!}{\left(\frac{n}{2}\right)!\,2^{\left(\frac{n}{2}\right)}}$$ ways to arrange all $n$ people into sets of pairs.

So for 8 people there are $\frac{8!}{4!\,2^{4}}=105$ possible sets of pairs.

Now the question remains, how many sets are there that do not contain any pairs from the original set?

Let $$f(x)=\frac{x!}{\left(\frac{x}{2}\right)!\,2^{\frac{x}{2}}}$$

If $S=\left\{p_{1}, p_{2}, ..., p_{\frac{n}{2}}\right\}$ is our original set of pairs and $P_{k}$ is the set of all sets containing $p_{k}$, where $1\le k\le\frac{n}{2}$ then by the inclusion-exclusion principal the number of sets not containing any of the original pair is:

$$f(n)-\left(|P_{1}|+|P_{2}|+\dots+|P_{\frac{n}{2}}|\right)+\left(|P_{1}\cap P_{2}|+(|P_{1}\cap P_{3}|+\dots\right)-\dots$$

But for any $P_{k}$; $|P_{k}|=f(n-2)$ therefore, $|P_{1}|=|P_{2}|=\dots=|P_{\frac{n}{2}}|$and in general, given any $k$ where $1\le k\le \frac{n}{2}$, then $|P_{1}\cap P_{2}\cap\dots\cap P_{k}|=f(n-2k)-\dots$

So the number of sets not containing any of the original pair is:

$$f(n)-\left(f(n-2)+f(n-2)+\dots\right)+\left(f(n-4)+f(n-4)+\dots\right)-\dots$$

which equals:

$$f(n)-{\frac{n}{2}\choose 1}f(n-2)+{\frac{n}{2}\choose 2}f(n-4)-\dots$$

So in the case where $n=8$

$$f(8)-4f(6)+6f(4)-4f(2)+1f(0)=105-60+18-4+1=60$$