How many ways are there to arrange 4 pairs of shoes in a row so that 2 shoes of the same pair are never next to each other?
I started by making a visual representation of the 8 shoes in a row in pairs,
shoes -> II II II II
Since there 4 pairs, I considered 4 conditions
$C_i$ = pair of shoes are never together (1 ≤ i ≤ 4)
N($\bar{C_1}\bar{C_2}\bar{C_3}\bar{C_4})$ = $S_0 - S_1 + S_3 - S_4$
$S_0$ = All possible arrangements = $\frac{8!}{(2!)^4}$ since there are 8 shoes, and 4 pairs are identical.
$S_1$ = One pair of shoes is identical = $4 \choose 1$$\frac{7!}{(2!)^3}$ Assuming, 2 shoes to be one pair, we are left with 7 shoes, and there's only 3 identical pairs left now
Following a similar pattern,
$S_2$ = $4 \choose 2$$\frac{6!}{(2!)^2}$
$S_3$ = $4 \choose 3$$\frac{5!}{(2!)}$
$S_4$ = $4 \choose 4$${4!}$
$N(\bar{C_1}\bar{C_2}\bar{C_3}\bar{C_4})$ = $\frac{8!}{(2!)^4}$ - $4 \choose 1$$\frac{7!}{(2!)^3}$ + $4 \choose 2$$\frac{6!}{(2!)^2}$ - $4 \choose 3$$\frac{5!}{(2!)}$ + $4 \choose 4$${4!}$
I'm not sure if what I did is correct, any suggestions would be appreciated.
