The first letter can be any one of 26. For each of these ...
The second letter can be any one of the remaining 25. For each of these ...
The third letter can be any one of the remaining 24.
So the number of different 3-letter line-ups is (26! / 23!) = (26 x 25 x 24) = 15,600.
That's the answer if you care about the sequence of the letters, i.e. if you call ABC and ACB different.
If you don't care about the order of the 3 letters ... if ABC, ACB, BAC, BCA, CAB, and CBA are all
the same to you, then there are six ways to arrange each group of 3 different letters.
Then the total number of different picks is [ 26! / (23! 6!) ] = (15,600/6) = 2,600 .
There are 26 different letters that can be chosen for each letter. There are 10 different numbers that can be chosen for each number. Since each of the numbers/digits that can be chosen for each of the six "spots" are independent events, we can multiply these combinations using the multiplicative rule of probability.combinations = (# of different digits) * (# of different digits) * (# of different digits) * (# of different letters) * (# of different letters) * (# of different letters) = 10 * 10 * 10 * 26 * 26 * 26 = 103 * 263 = 1000 * 17576 = 17,576,000 different combinations.
I assume you mean no repetition to be ABC and not ABA OR AAB. That being said this is very straight forward. The first position can potentially have any of the 26 letters. the second position can only have 1 of 25 possible letters because one letter has already be selected for the first position. And the final position can only have 1 of 24 possible letters because two letters have already been selected for the first two positions. Just multiply the number of possibilities for each position together and you have your answer. 26 * 25 * 24 = 15600 possible combinations.
720 is the number of ways to combine three known letters and three known numbers.For example, the letters A, B & C and the numbers 1, 2 & 3. The total combinations of these 6 characters is:(6 options)*(5 options)*(4)*(3)*(2)*(1) = 720.However, if the three numbers and three letters are unknown and any number or letter is possible, and repeated numbers or letters are acceptable (such as with a license plate), then the total possibilities for each "space" are multiplied together:(26 possibilities)*(26 possibilities)*(26 possibilities)*(10 possibilities)*(10 " ")*(10 " ") = 17,576,000 combinations.That is, there are 26 letters in the alphabet and 10 numbers (0 thru 9).This is assuming that three of the six spaces spaces are reserved for letters and three spaces are reservedfor numbers.If the combination can be any three letters and any three numbers where different combinations are made by changing whether each space contains a number or a letter, then the answer becomes a product and sum of different choose functions and is much more complicated...
26 x 25 x 24 = 15600
For every letter there are 26 possibilities, for every digit, 10. Multiply all of this together (26 x 26 x 10 x 10 x 10) = 676,000.
3,124,550 possible 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.
In California, for example, the first digit of a standard plate is a number, followed by 3 letters, and then three numbers. There are 26 letters in the alphabet, so there are 26 raised to the 3rd power combinations, or 26 * 26 * 26, which is 17,576 possibilities just of the 3 letters.
You could make 10*10*10*26*26*26 combinations, or 17576000 combinations.
Assuming you are using the standard English alphabet, the number of combinations you can make are: 26 x 26 = 676 combinations.
As there are 26 letters in the alphabet. You can calculate the number of combinations by multiplying 26x26x26, giving you the answer 17576.
If the first two letters must be an S or a T, then there are three remaining places that can be one of 26 letters. This is equal to 26 * 26 * 26 * 2, or 35,152 possible combinations.
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.
That depends - on how many letters are in the password - AND - if any letter can be repeated. For example... an eight-digit password has 26 x 26 x 26 x 26 x 26 x 26 x 26 x 26 = 208,827,064,576 possible combinations IF every letter can b repeated. If each letter can only be used once, then the formula is 26 x 25 x 24 x 23 x 22 x 21 x 20 x19 = 62,990,928,000 possible combinations.
There are 26 different letters that can be chosen for each letter. There are 10 different numbers that can be chosen for each number. Since each of the numbers/digits that can be chosen for each of the six "spots" are independent events, we can multiply these combinations using the multiplicative rule of probability.combinations = (# of different digits) * (# of different digits) * (# of different digits) * (# of different letters) * (# of different letters) * (# of different letters) = 10 * 10 * 10 * 26 * 26 * 26 = 103 * 263 = 1000 * 17576 = 17,576,000 different combinations.
When trying to work out how many different combinations there are, you need to know how many options there are for each value. If the password only contains lower case letters, then we have 26 options for each value. For each letter in the password, there are 26 options, so the total number of possible options is 26x26x26x26x26x26 or 266 This equals 308,915,776 so there are 308,915,776 possible different combinations of six letters.