How to match any combination of letters using regex?

Just to round out the collection:

^(?:([abc])(?!.*\1))+$

Want to handle a larger set of characters? No problem:

^(?:([abcdefgh])(?!.*\1))+$

EDIT: Apparently I misread the question; you're not validating individual strings like "abc" and "ba", you're trying to find whole-word matches in a larger string. Here's how I would do that:

\b(?:([abc])(?![abc]*\1))+\b

The tricky part is making sure the lookahead doesn't look beyond the end of the word that's currently being matched. For example, if I had left the lookahead as (?!.*\1), it would fail to match the abc in abc za because the lookahead would incorrectly flag the a in za as a duplicate of the a in abc. Allowing the lookahead to look only at valid characters ([abc]*) keeps it on a sufficiently short leash. And if there are invalid characters in the current word, it's not the lookahead's job to spot them anyway.

(Thanks to Honest Abe for bringing this back to my attention.)

Try this regex:

^([abc])((?!\1)([abc]))?((?!(\1|\2))([abc]))?$

Check in regexpal

^(?=([^a]*a?[^a]*)$)(?=([^b]*b?[^b]*)$)(?=([^c]*c?[^c]*)$)[abc]{1,3}$

This works with lookaheads.

It includes this pattern in three variations: (?=([^a]*a?[^a]*)$)

It says: There needs to be at most one a from here (the beginning) until the end.

Combining lookaheads and backreferences:

^([abc])((?!\1)([abc])((?!\1)(?!\3)[abc])?)?$

Use regex pattern

\b(?!\w*(\w)\w*\1)[abc]+\b

You can use this pattern with any set and size, just replace [abc] with desired set...

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Example:

(above output is from myregextester)