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Chapter 6: Problem 21

How many five-letter sequences are possible that use the letters \(\mathrm{b}, \mathrm{o}, \mathrm{g}, \mathrm{e}, \mathrm{y}\) once each?

Short Answer

Expert verified
There are 120 unique five-letter sequences that can be formed using the letters b, o, g, e, and y once each.

Step by step solution

01

Understanding the problem

We have 5 unique letters and would like to find the number of unique five-letter sequences that can be formed using these letters without repetition. This is a permutation problem.
02

Calculate the number of permutations

To arrange 5 unique objects into a sequence without repetition, we use the formula for the number of permutations: \[nPr = \frac{n!}{(n-r)!}\] Where n is the number of objects and r is the number of objects being arranged. In this case, n = 5 (5 distinct letters) and r = 5 (we are arranging all 5 letters).
03

Substitute the values and solve

Substituting the values into the formula: \[5P5 = \frac{5!}{(5-5)!}\] \[5P5 = \frac{5!}{0!}\] Since 0! = 1, \[5P5 = \frac{5!}{1}\] Now, calculate 5!: \[5! = 5\times4\times3\times2\times1 = 120\] So, the final result is: \[5P5 = \frac{120}{1} = 120\] Therefore, there are 120 unique five-letter sequences that can be formed using the letters b, o, g, e, and y once each.

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Most popular questions from this chapter

Use a Venn diagram or some other method to obtain a formula for \(n(A \cup B \cup C)\) in terms of \(n(\bar{A}), n(B), n(C)\), \(n(A \cap B), n(A \cap C), n(B \cap C)\) and \(n(A \cap B \cap C)\).

Let \(A=\\{H, T\\}\) be the set of outcomes when a coin is tossed, and let \(B=\\{1,2,3,4,5,6\\}\) be the set of outcomes when a die is rolled. Write each set in terms of A and/or \(B\) and list its elements. The set of outcomes when a die is rolled and then a coin tossed.

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