The word "DECAGON" has 7 letters, with the letter "A" appearing once, "C" appearing once, "D" appearing once, "E" appearing once, "G" appearing once, "N" appearing once, and "O" appearing once. To find the number of different 4-letter permutations, we need to consider combinations of these letters. Since all letters are unique, the number of 4-letter permutations is calculated using the formula for permutations of n distinct objects taken r at a time: ( P(n, r) = \frac{n!}{(n-r)!} ). Here, ( n = 7 ) and ( r = 4 ), so the number of permutations is ( P(7, 4) = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840 ). Thus, there are 840 different 4-letter permutations that can be formed from the letters in "DECAGON."
The letters from a to J consist of 10 distinct letters. The number of permutations of these letters is calculated using the factorial of the number of letters, which is 10!. Therefore, the total number of permutations is 10! = 3,628,800.
There are 9 * 8 * 7, or 504, three letter permutations that can be made from the letters in the work CLIPBOARD.
24
The answer can be written as "24!" or 24 factorial, which means 24 X 23 X 22 .... X 3 X 2 X1. The answer is 620,448,401,733,239,439,360,000 different combinations. --------------------------------------------------------------------------------------------- The question should say "permutations of 24 different letters" I believe, for which the above answer is. Number of 24 different letter permutations.
There are 7 factorial, or 5040 permutations of the letters in the word NUMBERS.
There are 8 letters in "geometry", so there are 8! (factorial) ways to arrange them in different permutations. 8! = 40,320 permutations.
If they are all different, then 40320.
There are 5*4*3 = 60 permutations.
360. There are 6 letters, so there are 6! (=720) different permutations of 6 letters. However, since the two 'o's are indistinguishable, it is necessary to divide the total number of permutations by the number of permutations of the letter 'o's - 2! = 2 Thus 6! ÷ 2! = 360
There are 75600 permutations.
3 x 8 x 4 = 96 different outfits, provided every possible combination is acceptable on the street.
The number of different ways the letters of a word can be arranged, when all the letters are different, is the same as the number of permutations of those letters. In this case, the answer is 5!, or 120.
Normally, there would be 5!=120 different permutations* of five letters. Since two of the letters are the same, we can each of these permutations will be duplicated once (with the matching letters switched). So there are only half as many, or 60 permutations.* (the correct terminology is "permutation". "combination" means something else.)
If you mean permutations of the letters in the word "obfuscation", the answer is 1,814,400.
There are 6! = 720 permutations.
There are 9 * 8 * 7, or 504, three letter permutations that can be made from the letters in the work CLIPBOARD.
The number of permutations of the letters SWIMMING is 8 factorial or 40,320. The number of distinct permutations, however, due to the duplication of the letters I and M is a factor of 4 less than that, or 10,080.