how many different ways to hit 21

[Blitzkrieg]

If you want 3 card combinations:
9,9,3: 3 combinations (993, 939, and 993)
9,8,4: 6 combinations (984, 948, 894, 849, 489, and 498)
9,7,5: 6 combinations
9,6,6: 3 combinations
8,8,5: 3 combinations
8,7,6: 6 combinations
7,7,7: 1 combination
Total number of combinations is 28.

Better yet, how many different combinations can make 21 using every possible combination of all 4 suits or less (2 cards is the least), with 52 cards to make 21?

28 seems inconclusive T3, as there are 4 suits that hold the value of a 7 pip card in a single deck. So there should be 4 different combinations of 21 using only the 7's in a single deck as I have listed below. Not 1 combination.

The OP didn't state the number of decks in use in his original question. However, as the number of decks increases to 2 decks, the number of different combinations totaling 21 increases accounting for only the 7. If there are 3 decks in play, then there is the possible combination of 3 similar 7's being dealt such as 7s, 7s, 7s. In 3 decks what is the frequency that any player would be dealt 3-of-a-kind suited on the first 3 cards of any particular hand before the cut card with one player at the table, even if the final hit was a possible suited bust card?

7h, 7d, 7s red-red-black
7h, 7s, 7c red-black-black
7d, 7c, 7s red-black-black
7d, 7h, 7c red-red-black