number of ways of arranging four pairs of socks in a line so that no adjacent socks form a pair 
Suppose there are four pairs of socks, let's label them $AA,BB,CC,DD$. The left and right socks of the pair are indistinguishable. How many ways are there to arrange them in a line such that no adjacent socks are a pair, e.g. $ABABCDCD$. In extension, how about if there are $n$ pairs? 

My attempt is as follows, but I feel it is too complicated. I am seeking an easier method. 
I tried to partition into four distinct cases:
$$ABCDXXXX, 432$$
$$ABACXXXX,120$$
$$ABCAXXXX,288$$
$$ABABXXXX, 24$$
So in total there are $864$ ways. This is the correct answer but it is complicated. 
 A: An alternative approach, which seems like it should generalise much more easily, is to use inclusion-exclusion.
There are $\dfrac{8!}{2!^4}$ ways to arrange the socks without restriction.
There are $\dfrac{7!}{2!^3}$ ways to arrange them with AA together (think of this as arranging the $7$ items AA,B,B,C,C,D,D).
There are $\dfrac{6!}{2!^2}$ ways to arrange them with AA together and BB together...
A: We can use the Inclusion-Exclusion Principle.
We have eight positions to fill.  We can fill two of them with A's in $\binom{8}{2}$ ways, two of the remaining six positions with B's in $\binom{6}{2}$ ways, two of the remaining four positions with C's in $\binom{4}{2}$ ways, and fill the final two positions with D's in $\binom{2}{2}$ ways.  Hence, there are
$$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
distinguishable arrangements of four pairs of socks.
From these, we must subtract those arrangements in which two adjacent socks form a pair.
Two adjacent socks form a pair:  There are $\binom{4}{1}$ ways to choose which of the four pairs of socks form an adjacent pair.  Say the two socks labeled A are adjacent.  Then we have seven objects to arrange: AA, B, B, C, C, D, D.  Choose one of the seven positions for the pair AA, two of the remaining six positions for the B's, two of the remaining four positions for the C's, then fill the final two positions with the D's.  The socks can be arranged in
$$\binom{7}{1}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
ways.  Hence, there are
$$\binom{4}{1}\binom{7}{1}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
arrangements in which two adjacent socks form a pair.
However, if we subtract this from the total, we will have subtracted too much since we will have subtracted each case in which there are two pairs in which two adjacent socks form a pair twice, once for each way we could have designated one of those pairs as the pair of adjacent socks that form a pair.  We only want to subtract them once, so we must add them back.
Two pairs in which two adjacent socks form a pair:  There are $\binom{4}{2}$ ways to select two pairs of socks.  Say the socks labeled A are adjacent and the socks labeled B are adjacent.  Then we have six objects to arrange:  AA, BB, C, C, D, D. They can be arranged in
$$\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$
ways.  Hence, there are
$$\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$
in which there are two pairs in which two adjacent socks form a pair.
If we add this to our running total, our answer will be too large.  This is because we first added each case in which there are three pairs in which adjacent socks form a pair three times, once for each way of designating one of those three pairs as the pair of adjacent socks that form a pair of socks, and subtracted them three times, once for each of the $\binom{3}{2}$ ways we could designate two of those three pairs as the pairs of adjacent socks that form a pair of socks.  Thus, we have not subtracted cases in which there are three pairs of socks in which adjacent pairs of socks form a pair.
Three pairs in which two adjacent socks form a pair:  There are $\binom{4}{3}$ ways of selecting the three pairs of socks.  Say the two socks labeled A are adjacent, the two socks labeled B are adjacent, and the two socks labeled C are adjacent.  Then we have five objects to arrange: AA, BB, CC, D, D.  They can be arranged in
$$\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{2}{2}$$
ways, so there are
$$\binom{4}{3}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{2}{2}$$
arrangements in which there are three pairs in which two adjacent socks form a pair of socks.
If we subtract this amount from our total, our count will be too small.  This is because we have counted each case in which there are four pairs in which adjacent socks form a pair of socks twice.  We subtracted them four times, once for each way we could designate one of those four pairs as the pair of adjacent socks that form a pair; added them six times, once for each of the $\binom{4}{2}$ ways we could designate two of those four pairs as the pairs of adjacent socks that form a pair of socks; then subtracted them four times, once for each of the $\binom{4}{3}$ ways we could designate three of those four pairs as the pairs of adjacent socks that form a pair of socks.  We only want to subtract them once, so we must add them back.
Four pairs of adjacent socks in which adjacent socks form a pair:  We have four objects to arrange: AA, BB, CC, DD.  They can be arranged in $4!$ ways.
By the Inclusion-Exclusion Principle, the number of admissible arrangements is
$$\binom{8}{2}\binom{6}{2}\binom{4}{2} - \binom{4}{1}\binom{7}{1}\binom{6}{2}\binom{4}{2} + \binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2} - \binom{4}{3}\binom{5}{1}\binom{4}{1}\binom{3}{1} + 4!$$
