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The hydrocarbon naphthalene has ten carbon atoms arranged in a double hexagon, and eight hydrogen atoms attached at each of the corners of the hexagons. Naphthol is obtained by replacing one of the hydrogen atoms of naphthalene with a hydroxyl group ($\mathrm{OH}$). How many isomers of naphthol are there?

So far I considered $N=\{1,2,3,4,5,6,7,8\}$ the set of hydrogen and $C=\{\text{hydrogen},\text{hydroxyl}\}$ the set of types. Considering hydrogen $\mathrm{H}$ and hydroxyl $\mathrm{D}$, I got, following an example, that $ZGv(\mathrm{H+D},\dots,\mathrm{H^8+D^8})$. No idea how to get the number of isomers out of this. The example I followed was at this link at page 40.

Is it the same thing as the cyclobutane from 6.6 on page 40, or does it come differently? It has two types, so it should be the same?

Tetramethylnaphthalene is obtained by replacing four of the hydro- gen atoms of naphthalene with methyl groups ($\mathrm{CH_3}$). How many isomers of tetramethylnaphthalene are there?

This is another one that implies the use of the same formula, is it solved on the same principle like the one above? Appreciate the help.

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First draw your compound leaving all hydrogen atoms blank for the moment. These will be our slots, to be filled with a hydrogen atom or a hydroxyl group. The base generating function is $1+z$, with $z$ marking the hydroxyl group.

The slots are being permuted by the automorphism group $G$ of the compound, which is very simple in structure. It contains four permutations: the identity, the horizontal flip, the vertical flip, and a rotation by 180 degrees. The cycle index of $G$ can then be read off the diagram and it is $ Z(G) = \frac{1}{4} \left( a_1^8 + 3 a_2^4 \right).$

Now do the usual substitution to obtain the orbit enumerator, which is $1+2\,z+10\,{z}^{2}+14\,{z}^{3}+22\,{z}^{4}+14\,{z}^{5}+10\,{z}^{6}+2\, {z}^{7}+{z}^{8}$

It follows that there are two isomers.

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    Math.stackexchange.com says no extended discussions -- so I'll adhere to that. The permutations are [1][2][3][4], [1,4][2,3], [1,3][2,4], [1,2] [3,4].2012-11-15