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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.

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    The left and right shoes in a pair are distinguishable. – 2017-01-19
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    I suppose you can find a few posts about this problem on this site, it is often formulated using married couples. For example, a question [about four couples](http://math.stackexchange.com/questions/68541/probability-question-about-married-couples) and a question [about three couples](http://math.stackexchange.com/questions/837072/traditional-combination-problem-with-married-couples-buying-seats-to-a-concert). – 2017-01-19
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    BTW if the main point of your question is to ask for checking and criticism of your solution (as opposed to asking for any solution from others), you should add the tag ([tag:solution-verification]) to indicate this. See the [tag-info](http://math.stackexchange.com/tags/solution-verification/info) for more details. – 2017-01-19

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