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Our teacher asked us how many geometrical isomers are possible with the formula $\ce{c2ClBrIf}$. I answered with the logic $^4C_2=6$ and it turned out to be correct.

Is the use of combinations this way generally helpful in determining the number of possible isomers? For example, if I have a rigid molecule with three carbon atoms and six substituents (perhaps a cyclopropane), is the correct number of possible isomers 20?

$$^6 C_3 =\dfrac{6!}{(6-3)!\cdot 3!}= 20$$

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    $\begingroup$ LOL - The reason?!? - You got lucky since you didn't understand the concept of isomers. $\endgroup$ – MaxW Dec 17 '15 at 19:29
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    $\begingroup$ @MaxW - Actually, the approach described in your answer is equivalent to the statement $^4C_2$, which is the number of distinct combinations of a set of four taken in pairs - in this case a set of four substituents divided over two carbon atoms. It is evaluated equivalently mathematically as well: $$^k C_j = \dfrac{k!}{(k-j)!j!}$$ $4!=24$, and you determined that only one in four structures is a different isomer (as every isomer can be drawn in four degenerate ways), and $(4-2)!\cdot 2!=4$. $\frac{24}{4}=6$ $\endgroup$ – Ben Norris Dec 17 '15 at 21:51
  • $\begingroup$ @BenNorris - Yes and no. You need to look for the general principle not just a particular example. // Your edit greatly changed and expanded the question... $\endgroup$ – MaxW Dec 17 '15 at 22:10
  • $\begingroup$ What is $\ce{c2ClBrIf}$? Do you mean $\ce{C2ClBrIF}$? $\endgroup$ – Martin - マーチン Dec 18 '15 at 17:03
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For the following I've numbered the positions on ethylene where the four halogen atoms could go. (The Carbon-Carbon double bond doesn't "spin.") It doesn't matter which halogen goes to 1, so let's make it Fluorine. If Fluorine were in any other position then you could flip or rotate the molecule to get Fluorine to position 1. Now there are three choices left for position 2, which leaves 2 choices for position 3. So the number of possible isomers is:

$3 \times 2 = 6$

Note that if you just willy nilly drew molecules then you could draw 4! or 24 permutations. However when you rotate and flip the molecules, then only 6 unique molecules exist.

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  • $\begingroup$ Okay I get what MaxW was trying to explain there. But here with another logic can we just not take in account the case of circular permutation and get the total number of geometrical isomers as 3!=6? $\endgroup$ – user34221 Aug 28 '16 at 1:45

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