6
Assuming you can have duplicate letters such as AAA, AAB etc... there are 26 x 26 x 26 combinations - which is a total of 17,576 permutations !
There are 3 consonants from 21 and 2 vowels from 5. That gives 21C3 * 5C2 combinations = 21*20*19/(3*2*1) *5*4/(2*1) = 1330*10 = 13300 combinations in all.
233 = 18 combinations.
3 items each in 3 categories gives 3*3*3 = 27 possible combinations.
You did not give us the set to work with...
6
There are: 35C3 = 6545 combinations
Assuming you can have duplicate letters such as AAA, AAB etc... there are 26 x 26 x 26 combinations - which is a total of 17,576 permutations !
Oh, dude, there are like 26 letters in the alphabet, right? So, for each position in a 3-letter combination, you have 26 choices. That means you'd have 26 choices for the first letter, 26 for the second, and 26 for the third. So, the total number of 3-letter combinations would be 26 x 26 x 26, which is... math.
There are 3 consonants from 21 and 2 vowels from 5. That gives 21C3 * 5C2 combinations = 21*20*19/(3*2*1) *5*4/(2*1) = 1330*10 = 13300 combinations in all.
233 = 18 combinations.
There are 6C3 = 20 such combinations.
There are 5C3 = 10 combinations.
256 iThink * * * * * It depends on combinations of how many. There is 1 combination of 4 letters out of 4, 4 combinations of 3 letters out of 4, 6 combinations of 2 letters out of 4, 4 combinations of 1 letter out of 4. Than makes 15 (= 24-1) in all. Well below the 256 suggested by the previous answer.
3 items each in 3 categories gives 3*3*3 = 27 possible combinations.
11*3*3 = 99 combinations.