The letters ABC can be arranged in 3! = 3 x 2 x 1 = 6 ways. This is because there are 3 letters to arrange, and for each position, there are 3 choices for the first letter, 2 choices for the second letter, and 1 choice for the last letter. Therefore, the total number of ways to arrange the letters ABC is 6.
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You have 3 letters to be put into 3 spaces. You have 3 ways of choosing the first letter followed by only 2 ways of choosing the second, leaving only 1 way to place the third. So the number of ways of arranging the 3 letters is 3 x 2 x 1 = 6
The first letter can be any one of 22. For each of these ...The second letter can be any one of the remaining 21. For each of these ...The third letter can be any one of the remaining 20.So the number of different 3-letter line-ups is (22 x 21 x 20) = 9,240.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 (9,240/6) = 1,540.
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The number of 3 letter different combinations that can be made from 10 differentletters is given by: 10C3 = 10!/[3!(10-3)!] = 120different 3 letter combinations.The number of three letter different permutations (one given 3 letter combinationgives 3! = 6 different permutations, e.g., the combination abc gives the 9permutations, abc, acb, bac, cab, bca, cba) is given by 10P3 = 10!/(10-3)! = 720different 3 letter permutations.The symbol (!) is the operator called "factorial" and it implies the multiplication:n! = n∙(n-1)! = n∙(n-1)∙(n-2)! = n(n-1)(n-2)(n-3)...(4)(3)(2)(1)(0!)where 0! = 1.i.g., 6! = 6∙5∙4∙3∙2∙1 = 720