# Use of combinations to determine number of isomers

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?

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

• LOL - The reason?!? - You got lucky since you didn't understand the concept of isomers.
– MaxW
Dec 17 '15 at 19:29
• @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 $(4-2)!\cdot 2!=4$. $\frac{24}{4}=6$ Dec 17 '15 at 21:51
• @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...
– MaxW
Dec 17 '15 at 22:10
• What is $\ce{c2ClBrIf}$? Do you mean $\ce{C2ClBrIF}$? Dec 18 '15 at 17:03

$3 \times 2 = 6$ 