Three 7-letter words, 21 unique letters, excludes either HM or QS

Question

I'm looking for three seven letter words consisting of 21 unique letters, and the final result must exclude either the letters HM or QS (Excluding S may be impossible based off of my own attempts). One word MUST include V. The closest I was able to get by hand (admittedly without the excluded letters) is:

• Swarthy Flocked Bumping (or Jumping)
• Mitzvah Cyborgs Flunked (or Plunked)

Please use the YAWL wordlist.

Answer 1

Here are some found solutions, there should be all of them in the given word list.

#     WORD1     WORD2     WORD3    UNUSED
-----------------------------------------
1    BRAZING + KVETCHY + MUDFLOW    JPQSX
2    FLIGHTY + PROVEND + ZAMBUCK    JQSWX
3    PERVING + TWYFOLD + ZAMBUCK    HJQSX
4    PREVING + TWYFOLD + ZAMBUCK    HJQSX
5    PROVEND + WIGHTLY + ZAMBUCK    FJQSX

Five solutions only...

@Lukas Rotter Many thanks for pointing out the error in the previous version!

---

Used newly UPDATED py code:

ALL_LETTERS = set('ABCDEFGHIJKLMNOPQRSTUVWXYZ')

words = []
with open('word.list') as stream:
    for line in stream.readlines():
        word = line.strip()
        if len(word) == 7 and len(set(word)) == 7:
            words.append(word.upper())

print(len(words))

vowels = set('AEIOUY')
dic = {}    # vowels --> list of words involving these vowels

for word in words:
    used_vowels = list(vowels.intersection(word.upper()))
    used_vowels.sort()
    used_vowels = ''.join(used_vowels)

    if used_vowels not in dic:
        dic[used_vowels] = []
        
    dic[used_vowels].append(word)
    
keys = list(dic)
keys.sort()

solutions = []
for ae in keys:
    for io in keys:
        if io <= ae:    continue 
        if set(ae).intersection(io):    continue
        for uy in keys:
            if uy <= io:    continue
            if set(uy).intersection(ae + io):    continue

            # print(f"TRYING {ae} {io} {uy}")
            for ae_word in dic[ae]:
                for io_word in dic[io]:
                    if len(set(ae_word + io_word)) < 14:    continue
                    for uy_word in dic[uy]:
                        letters = ae_word + io_word + uy_word
                        if len(set(letters)) < 21:    continue

                        unused = list(ALL_LETTERS.difference(letters))
                        unused . sort()
                        if ( 'V' in letters
                             and ( set('HM').issubset(unused) or 
                                   set('QS').issubset(unused) ) ):

                            sol = [ae_word, io_word, uy_word]
                            sol.sort()

                            solutions.append((sol, ''.join(unused)))

solutions.sort()
for counter, (sol, unused) in enumerate(solutions):
    ae_word, io_word, uy_word = sol
    print(f'    {counter + 1}    {ae_word} + {io_word} + {uy_word}    {unused}')
                  

(The code was rearranged in the final check condition to be simpler to digest also for readers that have no or limited python experience.)

Answer 2

The MVP here is the St John Ambulance attendant

brazing, kvetchy, mudflow
flighty, provend, zambuck
perving, twyfold, zambuck
preving, twyfold, zambuck
provend, wightly, zambuck