Well, if you don't mind doing a bit of extra work, you can easily try all 255 nontrivial subsets of 7 tiles by cycling a counter from 1 to 255, and using a bitmask to check on each increment whether bit n is 1 (and thus whether tile n should be included).

Of course, this doesn't take into account that two tiles could be the same letter. You may be able to save a bit of effort if you figure out how to deal with that (non-trivial).