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\begin{tabular}{l} Answer the questions below. \\ \( \begin{array}{l}\text { (a) There are } 15 \text { European cities that Juan would eventually like to visit. On his next vacation, } \\ \text { though, he only has time to visit } 4 \text { of the cities: one on Monday, one on Tuesday, one on Wednesday, } \\ \text { and one on Thursday. He is now trying to make a schedule of which city he'll visit on which day. How } \\ \text { many different schedules are possible? (Assume that he will not visit a city more than once.) } \\ \square\end{array} \begin{array}{l}\text { (b) To log on to a certain computer account, the user must type in a 4-letter password. In such a } \\ \text { password, no letter may be repeated, and only the lower case of a letter may be used. How many } \\ \text { such 4-letter passwords are possible? (There are } 26 \text { letters in the alphabet.) } \\ \square\end{array} \) \\ \hline\end{tabular}

Ask by Robbins Stuart. in the United States
Jan 06,2025

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There are 32,760 different schedules possible for visiting the cities and 358,800 possible 4-letter passwords.

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To find how many schedules Juan can create, we’ll need to use permutations since the order of the cities matters. For the first day, he has 15 options, for the second day he’ll have 14 options (as he cannot visit the same city), for the third day there’ll be 13 options, and for the fourth, 12 options left. Therefore, the total number of different schedules is calculated as \(15 \times 14 \times 13 \times 12\), which equals 32,760 unique schedules! Now for the passwords: since no letters can repeat, it’s a straightforward permutation problem again! The first letter has 26 options, the second will have 25 options left, the third will have 24, and the fourth will have 23. Therefore, the number of different 4-letter passwords is \(26 \times 25 \times 24 \times 23\), which calculates to a staggering 358,800 possible passwords!

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