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.
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.
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
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 allthe 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 .
dictinct object or letters- It implies that each object or letters differs in some way from the every other object or letters in the set Ex. Banana=B,a,n distinct letters
It is 720.
How many three-letter "words" can be made from 10 letters "FGHIJKLMNO" if repetition of letters are not allowed
That repetition would involve an onomatopoeia.
3,124,550 possible combinations
subdermatoglyphic
6
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.
Only 5 ways: one each with one of the letters missing.
12
The number of 3-letter codes that can be formed without repetition is equal to the number of choices for the first letter (4 letters), multiplied by the number of choices for the second letter (3 letters), multiplied by the number of choices for the third letter (2 letters). Therefore, the total number of 3-letter codes that can be formed is 4 x 3 x 2 = 24.
Four combinations: tam ham mat hat Hope it helps (:!
No is a word. On is a word.