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.
4*3*2*1 = 24 different combinations.
Just 4: 123, 124, 134 and 234. The order of the numbers does not matter with combinations. If it does, then they are permutations, not combinations.
As there are 26 letters in the alphabet. You can calculate the number of combinations by multiplying 26x26x26, giving you the answer 17576.
Six combinations: 123, 132, 213, 231, 312, 321
Taking this bit by bit... For each number there are 10 possibilities so three numbers has 10^3 which is 1000 for each letter there are 26 choices, so for three letters you have 26^3 which is 17576 together there are 17576 * 1000 options which makes 17576000 possible combinations of letters and numbers and I suspect that is what you want. However if the order of the combination is important you must take into account the position of each letter/number and remove duplicate combinations. Good luck with that
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 !
3,124,550 possible combinations
27^27
233 = 18 combinations.
There are 5C3 = 10 combinations.
There are 6C3 = 20 such combinations.
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.
3 items each in 3 categories gives 3*3*3 = 27 possible 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.