If the order matters: 10 x 9 x 8 (i.e., there are 10 options for choosing the first letter, 9 options for the second, 8 for the third one).
If the order doesn't matter, devide the previous result by 3! = 1 x 2 x 3.
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Starting with three different letters, six two-letter combinations can be made, if the order of the two letters is important. Only three combinations can be made if the order of the two letters is not important. Example: ABC AB AC BA BC CA CB - six variations But if (for your purposes) BA is the same as AB, Then there are only three: AB AC BC
Four combinations: tam ham mat hat Hope it helps (:!
There are 26 letters in the alphabet.If letters may be repeated in the airport code, then there are (26 x 26 x 26) = 17,576 possibilities.If adjacent letters can't be duplicates, then there are (26 x 25 x 25) = 16,250 possibilities.If all three letters must be different, then there are (26 x 25 x 24) = 15,600 possibilities.
Because there are 7 letters that don't repeat, 7X6X5X4=840 combinations. To give you an idea of how to solve the problem, if there had been 9 letters and you wanted to make combinations of 3 letters, you would multiply: 9X8X7=504