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
It is 720.
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.
This would be 26 x 26 x 10 x 10 = 67,600. Provided repetition is allowed.
4*4*4 = 64 ways.
10 (x 6 if no repetition is allowed but abc is not regarded as the same as acb), 125 if it is. ABC/ABD/ABE/ACD/ACE/ADE/BCD/BCE/BDE/CDE
How many three-letter "words" can be made from 10 letters "FGHIJKLMNO" if repetition of letters are not allowed
many i believe 8 or so??
There are 676,000 ways to make the license plates.
It is 720.
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.
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.
Assuming no repetition, 12 x 11 x 10.... x 4 x 3 x 2. If repetition is allowed, 1212
Alliteration is the repetition of the same consonant sounds and assonance is the repetition of the same vowel sounds.
It is (5/26)4 = 625/456,976 = 0.00137 approx.
That repetition would involve an onomatopoeia.
uncopyrightable
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.)