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Start with the five vowels; a, e, i, o, u and pick the seven most used consonants; t, n, s, h, r, d, l. Then, realize, that you can only make 12-letter words with 5 vowels and 7 consonants!

However, assuming that you meant from a pool of 5 vowels and 7 consonants, and assuming that you can use each letter only once, there would be a total of 12 X 11 X 10 = 1320 possible words, which I will abstractly start calling "3-letter combinations," since many of the 3-letter combinations I calculate won't actually end-up being words.

Now, assuming at least one of those letters has to be a vowel, there would be 7 X 6 X 5 (3-letter combinations with 2 consonants and 1 vowel of the form C-C-V) + 7 X 5 X 6 (3-letter combinations with 2 consonants and 1 vowel of the form C-V-C) + 5 X 7 X 6 (3-letter combinations with 2 consonants and 1 vowel of the form V-C-C) + 7 X 5 X 4 (3-letter combinations with 1 consonant and 2 vowels of the form C-V-V) + 5 X 7 X 4 (3-letter combinations with 1 consonant and 2 vowels of the form V-C-V) + 5 X 4 X 7 (3-letter combinations with 1 consonant and 2 vowels of the form V-V-C) + 5 X 4 X 3 (3-letter combinations with 0 consonants and 3 vowels of the form V-V-V) = 210 + 210 + 210 + 140 + 140 + 140 + 60 = 1110 possible, legal 3-letter combinations.

A double check of the above math is to realize that the only disallowed 3-letter combinations are those having 3 consonants, of which there are:

7 X 6 X 5 = 210.

Thus, the allowed number of 3-letter combinations plus the disallowed number of 3-letter combinations should equal the total possible number of 3-letter combinations, which was calculated above:

1110 + 210 = 1320, which checks out.

If "y" were allowed to take the spot of a vowel, then you would be able to additionally make (6 X 5 X 1) + (6 X 1 X 5) + (1 X 6 X 5) = 30 + 30 + 30 = 90 3-letter combinations, bringing the maximum amount of possible, legal 3-letter combinations to 1200.

If you were allowed to repeat the same letter within a 3-letter combination, the math changes to (7 X 7 X 5) X 3 + (7 X 5 X 5) X 3 + (5 X 5 X 5) = 735 + 525 + 125 = 1385 possible, legal 3-letter combinations, plus an additional (6 X 6 X 1) X 3 + (6 X 1 X 1) X 3 + (1 X 1 X 1) = 108 + 18 + 1 = 127 3-letter combinations if "y" could be a vowel (the 1 X 1 X 1 3-letter combination in this case is "yyy")

This brings the total to 1385 + 127 = 1512 possible, legal 3-letter combinations out of 12 X 12 X 12 = 1728 possible 3-letter combinations.

How many of these 3-letter combinations are actual words in the English language? That depends on the consonants chosen and the set of rules agreed upon. However, there are far to many options to go over in my lifetime, so if I were you, I'd start by using the initial set of letters I wrote down at the beginning and using the first set of rules layed out; i.e. you can't repeat consonants and "y" isn't a vowel, which is a gimme in this case since "y" isn't one of the 7 most-used consonants in the English language. That would mean you'd only have 1110 possible combinations to look at and judge to see if they were words or not.

It turns out that there are 198, 3-letter combinations of the 12 letters I wrote at the beginning that form words, at least according to a website that I linked below:

ADO ADS AID AIL AIN AIR AIS AIT ALE ALS ALT AND ANE ANI ANT ARE ARS ART ASH ATE DAH DAL DEL DEN DIE DIN DIS DIT DOE DOL DON DOR DOS DOT DUE DUI DUN DUO EAR EAT EAU EDH ELD ELS END ENS EON ERA ERN ERS ETA ETH HAD HAE HAO HAS HAT HEN HER HES HET HID HIE HIN HIS HIT HOD HOE HON HOT HUE HUN HUT IDS INS ION IRE ITS LAD LAR LAS LAT LEA LED LEI LET LEU LID LIE LIN LIS LIT LOT NAE NAH NET NIL NIT NOD NOH NOR NOS NOT NTH NUS NUT OAR OAT ODE ODS OES OHS OIL OLD OLE ONE ONS ORA ORE ORS ORT OSE OUD OUR OUT RAD RAH RAN RAS RAT RED REI RES RET RHO RIA RID RIN ROD ROE ROT RUE RUN RUT SAD SAE SAL SAT SAU SEA SEI SEL SEN SER SET SHA SHE SIN SIR SIT SOD SOL SON SOT SOU SRI SUE SUN TAD TAE TAN TAO TAR TAS TAU TEA TED TEL TEN THE THO TIE TIL TIN TIS TOD TOE TON TOR TUI TUN UDO UNS URD URN USE UTA UTS

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Q: How many 3 letter words can you make with 5 vowels and 7 consonants?
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