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4 votes
Aima got a chance for a free dinner at a restaurant from June 1 to June 10, 2023. Aima can choose more than one arrival date during this period as long as the dates are not consecutive. If Aima plans to come at least once, the number of possible arrival schedules that Aima can make is...

A. 45 B. 143 C. 144 D. 2025

2 Answers

3 votes

Answer:

c.144

Explanation:

answered
User Tobbe
by
8.0k points
5 votes

Answer: the answer is C 144

Step-by-step explanation:The number of possible arrival schedules that Aima can make can be determined by considering the number of ways she can choose her arrival dates during the given period. Since Aima cannot choose consecutive dates, she must choose alternate dates. Let's break down the problem step by step: 1. Aima can choose any arrival date from June 1 to June 10 as her first arrival. 2. For her second arrival, she can choose any date except the one immediately following her first arrival. 3. Similarly, for her third arrival, she can choose any date except the one immediately following her second arrival. 4. This pattern continues until her last arrival. To find the number of possible arrival schedules, we can use a concept called "stars and bars" or combinations. In this case, the "stars" represent the total number of arrival dates Aima can choose from (which is 10), and the "bars" represent the number of gaps between the chosen dates. Since Aima can choose more than one arrival date, we can have up to 9 gaps between the chosen dates. Therefore, we have 9 "bars" to distribute among the 10 "stars". This can be calculated using the combination formula. The number of possible arrival schedules is given by C(10+9, 9) = C(19, 9), which is equal to 144. Therefore, the correct answer is C. 144.


hope this helps ☀️

answered
User Vetsin
by
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