Given a list of words which contains the letters a-z at least once, how would you write a program to find the shortest pangram counted by number of characters (not counting spaces) as a combination of the words?

Since I am not sure whether short answers exist, this is not code golf, but rather just a discussion of how you would approach this. However, if you think you can manage to write a short program that would do this, then go ahead, and this might turn into code golf :)

I would approach this by proving that the problem is NP-hard, and by checking heuristics for the NP-hard problems that look similar.

We can reduce a Set Cover problem to our one. Set Cover is different in that not a number of letters used is minimized, but a number of words used is minimized instead. Assume we want to solve Set Cover problem, given N words, each of length less than M. Let's build another set of words by cloning the given set, but concatenating to each of them N*M non-english letters, say, Ж. If we could build a pangram (over a,b,c...x,y,z,ж alphabet) that requires minimum symbols, that would be a pangram with minimum words, if we remove all Ж letters.

This proves that the original problem is NP-hard, but, unfortunately we need a reduction *to* some NP-hard problem to reuse its (hopefully already known) heuristic. Set-Cover has a greedy heuristic with logarithmic approximation, but I don't think it applies to the original problem (nature of the Set-Cover problem requires taking letter-rich, long words; it's not the way to solve our problem).

So I'd search a list of related NP-hard problems, and check if there's something of interest. That's how I'd approach this one.