Wiki User
β 9y agoAssuming the mathematical definition of "combination" (ie unordered sets) and the standard 26 letter Latin Alphabet as used by English:
There are 5 vowels and 21 consonants, giving:
For each vowel: a choice of 3 consonants from 21, giving 5 x 21 C 3 = 5 x (21x20x19)/(3x2x1) = 6650 combinations
[Other languages have different size alphabets and different vowel/consonants (eg Welsh, which uses the Latin Alphabet characters, has 7 vowels and 23 consonants.)]
If you have a subset of the full Latin Alphabet, or another Alphabet, then you'll need to supply the set of letters for which you wish an answer.
Wiki User
β 9y agoWiki User
β 9y agoIf you want to ask questions about the "this set", then I suggest that you make sure that the question does contain some information about the set.
The number of combinations of 6 letters is 6! or 720.
7,893,600 (seven million, 8 hundred ninety-three thousand, 600) combinations in English.
432
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...
There is no word that can consist of 7 consonants and only have 3 letters. Consonants are letters.
There are 247 letters in the Tamil alphabet, including combinations of vowels and consonants.
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.
In the phrase "bouquet of flowers," there are 12 letters. Out of these 12 letters, 7 are consonants (excluding the spaces). So the fraction of consonants in the phrase is 7/12.
Consonants are the letters of the alphabet that are not vowels.
The word has four letters. All except "a" are consonants, or 3 out of 4. That means that 3/4 of the letters in that word are consonants.
No, it is not an adjective. Consonants (non-vowel letters) is a plural noun.
There are 2 consonants in the word "and" - the letters "n" and "d".
Tamil has 247 letters in it. 12 vowels, 18 consonants, 18x12 vowel-consonants and 1 special letter.
In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R. Number of ways of selecting 2 vowels out of 3 vowels = Number of ways of selecting 3 consonants out of 5 consonants = Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30 Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways. Hence, required number of different words = 30 × 5! = 3600
The question is invalid as consonants are individual letters that are not vowels . Therefore there are no consonants that end in es
Vowels and consonants are letters, not words. In the word "pick", the letter "i" is a vowel and the other three letters are consonants.