probability of pulling out non-matching socks one after another. 
“”So I was thinking that $P(\text {no two matching socks next to each other}) = 1 - P(\text {at least one matching socks next to each other}).$
Letting $\text{“sock $i$ together”}$ mean the left sock $i$ and right sock $i$ being together,
the above is equivalent to $1 - P(\text{sock 1 together} \cup \text{sock 2 together} \cup \cdots\cup \text{sock $n$ together})$, where $\cup$ is the union.
Then I could use principle of inclusion-exclusion to expand $P(\text{sock 1 together U sock 2 together} \cup\cdots\cup \text{sock $n$ together}) = P(\text{sock 1 together}) + \cdots + P(\text{sock $n$ together}) - (P(\text{sock 1 together $n$ sock 2 together}) + \cdots) + P(\text{sock 1 together $n$ sock 2 together $n$ sock 3 together}) +\cdots \text{etc.}$
This seems like it's going to get very long (esp with n different pairs), and the multiple intersections make the probabilities harder to calculate as well—so I was wondering if people could help me figure out a perhaps more concise way.
 A: Another way to answer this question that wasn't illustrated in the posts mentioned by @MathLover is by using a "balls in bins" counting strategy.
Suppose we've enumerated pairs of socks with tags $1,2,\ldots,n$ and we call $E_j$ the event that the $j^{\text{th}}$ pair of socks is together. In order to apply the principle of inclusion$-$exclusion, we must develop an expression for $P(E_1\cap \ldots \cap E_j)$ which is the probability that pairs $1,\ldots ,j$ are all together.
Let's say you're holding onto pairs $1,\ldots ,j$ while the remaining $2n-2j$ socks are laid out in front of you in an arbitrary order. Note there are $2n-2j+1$ spaces in which you can insert these $j$ pairs of socks. Using a "balls in bins" counting strategy, we see there are $j! \cdot A_j$ ways to insert the $j$ pairs into the $2n-2j+1$ empty spaces, where $A_j$ is the number of non$-$negative integer solutions to $x_1 + \dots + x_{2n-2j+1}=j$. From stars and bars, $A_j={2n-j \choose 2n-2j}$, and since the socks within each pair may be rearranged in $2$ different ways, we have that $$P(E_1 \cap \ldots \cap E_j)=\frac{{2n-j \choose 2n-2j}\cdot j!\cdot 2^j \cdot (2n-2j)!}{(2n)!}$$ So from inclusion$-$ exclusion, $$\begin{eqnarray*}P(E_1 \cup \ldots \cup E_n)&=&\sum_{j=1}^n (-1)^{j-1}{n \choose j}P(E_1 \cap \ldots \cap E_j) \\ &=& \frac{n!}{(2n)!}\sum_{j=1}^n \frac{(-1)^{j-1}\cdot (2n-j)!\cdot 2^j}{j! \cdot (n-j)!}\end{eqnarray*}$$ Your answer is $1-P(E_1 \cup \ldots \cup E_n)$.
