Final answer:
By substituting all hydrogen atoms in methane with chlorine or fluorine, there are 14 variations of CFCs that can be created after excluding cases with only fluorine or chlorine.
Step-by-step explanation:
The question concerns the number of variations of chlorofluorocarbons (CFCs) that can be created by substituting all hydrogen atoms in a methane molecule with chlorine or fluorine atoms. A methane molecule (CH4) has four hydrogen atoms that can be replaced.
If we consider each hydrogen atom being replaced independently by either chlorine or fluorine, there is a total of two choices (chlorine or fluorine) for each hydrogen atom.
Since there are four hydrogen atoms, we use the formula for permutations with repetition: 24 = 16. However, since we are interested only in substitutions that result in CFCs (which contain both chlorine and fluorine), we must subtract the variations that contain only fluorine and only chlorine, which are 2 (one for C4F4 and one for C4Cl4). Therefore, the correct answer is 16 - 2 = 14 variations of CFCs that can be created.