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How many three-letter words can be made from 10 letters FGHIJKLMNO if repetition of letters?
How many three-letter "words" can be made from 10 letters "FGHIJKLMNO" if repetition of letters are not allowed
How many 5 letters can be formed out of the letters of the word mathematics of repetition is not allowed?
There are eight (8) different letters in that word, so that's the number ofletter choices you have in each 5-letter group without repetition.The number of different 5-letter "words" is (8 x 7 x 6 x 5 x 4) = 6,720 .(They don't necessarily mean anything. They're just distinct sequencesof 5-letters each, like Morse-code random practice groups.)
If a license plate consists of two letters followed by four digits how many different plates are possible?
If all letters and numbers are allowed, the possibilities are 26x26x10x10x10x10. So: 6760000 different plates.
How many combinations of six letters are possible from six?
With 6 different letters, no repetition (like you have 6 different Scrabble tiles), and if order is important, then the proper statistical term is permutation, rather than combination, but many people refer to this type of application as a combination.The permutation (no repetition) of n items taken rat a time is n!/(n-r)! with in this case n=6 and r=6. The exclamation mark is notation for factorial.6! = 6 * 5 * 4 * 3 * 2 * 1. (6-6)! = 0! which is defined as 0! = 1So we have 6 * 5 * 4 * 3 * 2 * 1 = 720 different possibilities.
How many combinations of 3 different letters are there in the alphabet?
One way to work this out is to imagine that you have a bag containing all 26 letters of the alphabet.Take one letter from the bag. It could be any one of the 26, so there are 26 possibilities for your first letter.Take another letter from the bag. This time there are only 25 letters in the bag (you've already taken one out, remember!). So you have 25 possibilities for the second letter.So far there are 26 x 25 (= 650) possibilities for your first two letters. (AB, BA, AC, CA, AD, DA, etc)Now take your third letter from the bag. But remember, there are now only 24 letters remaining.So, for 'from 26 chose 3', altogether you have 26 x 25 x 24 possibilities, which comes to 15,600 different combinations of three different letters.
How many three-letter words can be made from 10 letters FGHIJKLMNO if repetition of letters?
How many three-letter "words" can be made from 10 letters "FGHIJKLMNO" if repetition of letters are not allowed
How many different three letter words can be made from the word ORANGE if repetition of letters is not allowed?
many i believe 8 or so??
What is the answer for the number of 3 letter words with or without meaning that can be formed out of the letters of the word LOGARITHMS if repetition of the letters is not allowed?
It is 720.
What 3-letter codes can you make using the letters A-J-L-W?
If repetition is allowed the formula would be 4x4x4 = 64 codes. If you must chose a different letter each time (no repetition) the formula would be 4x3x2 = 24 codes.
Find the number of ways in which the letters of the word SUM can be arranged in a row if repetition is allowed?
The number of permutations of the letter SUM is 3 factorial, or 6. Since none of those letters are repeated, the issue of repetition is not a factor. Perhaps the questioner meant to use a different word. If so, please restate the question.
What are some words containing the letters FF?
coffeeofferproffer
How many combination's can you get out of 12 numbers and letters?
Assuming no repetition, 12 x 11 x 10.... x 4 x 3 x 2. If repetition is allowed, 1212
What is a repetition of letters in a word called?
Alliteration is the repetition of the same consonant sounds and assonance is the repetition of the same vowel sounds.
Repetition of consonant letters or sounds?
That repetition would involve an onomatopoeia.
What is the probability that if 4 letters are typed with repetition allowed all 4 will be vowels?
It is (5/26)4 = 625/456,976 = 0.00137 approx.
Longest word without repetition of letters?
uncopyrightable
How many three-letter code symbols can be formed using a set of 5 different letters if repetition is not allowed?
First letter in code can be any of 5, second can be any of remaining 4 and so on. 5 x 4 x 3 x 2 = 120
