Solve the problem.
-How many different 8-letter codes are there if only the letters A, B, C, D, E, F, G, H, and I can be used and no letter can be used more than once?
A) 4,782,969
B) 36
C) 7
D) 181,440

Question

Quizplus · Accepted Answer

This problem involves calculating permutations because we are arranging a set number of items (letters) and order matters. Since there are 9 letters to choose from and we want to arrange 8 of them without repetition, we use the permutation formula P(n, r) = n! / (n-r)!, where n is the total number of items to choose from, r is the number of items to arrange, and "!" denotes factorial (the product of all positive integers up to that number). Here, n = 9 and r = 8, so P(9, 8) = 9! / (9-8)! = 9! / 1! = 9! = 362,880. However, the correct answer is not listed directly, indicating a mistake in my explanation. The correct calculation should be for arranging 8 letters out of 9, which indeed is 9! / (9-8)! = 9! / 1! = 9! = 362,880, but the correct answer choice provided is D) 181,440, which suggests a misunderstanding in the calculation process. The correct approach is to calculate the permutations of 8 out of 9 letters, which is 9P8 = 9! / (9-8)! = 9! / 1! = 362,880, not 181,440. Therefore, there seems to be a discrepancy in the interpretation of the answer choices.

Solve the Problem. -How Many Different 8-Letter Codes Are There If Only the If