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Chapter 8: Problem 63

Solve by the method of your choice. How many different four-letter passwords can be formed from the letters \(A, B, C, D, E, F,\) and \(G\) if no repetition of letters is allowed?

Short Answer

Expert verified
The total number of different four-letter passwords that can be formed from the given letters is \(7 * 6 * 5 * 4 = 840\).

Step by step solution

01

Understanding Permutations

Permutations refers to the arrangement of items in a specific order. In this context, it's about creating a password by arranging the seven given letters (A, B, C, D, E, F, G) in an order to form a unique four-letter password.
02

Calculating Permutations without Repetition

When repetitions are not allowed, the number of available choices reduces as each letter gets selected. For the first letter of the password, there are 7 options. Once that's chosen, there are 6 options left for the second letter. For the third letter, there are 5 options, and for the fourth and final letter, there are 4 options left. Thus, the total number of permutations is determined by multiplying the number of choices for each letter.
03

Performing the Calculation

The total number of permutations without repetition for a four-letter password from the seven given letters is calculated as follows: \(7 * 6 * 5 * 4\).

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Most popular questions from this chapter

Many graphing utilities have a sequence-graphing mode that plots. the terms of a sequence as points on a rectangular coordinate system. Consult your manual, if your graphing utility has this capability, use it to graph each of the sequences. What appears to be happening to the terms of each sequence as \(n\) gets larger? $$a_{n}-\frac{2 n^{2}+5 n-7}{n^{3}} n:[0,10,1] \text { by } a_{n}:[0,2,0.2]$$

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