Number of ways to schedule activities using combination or permutation.

Question

I'm trying to review for Probabilities and Statistics and came upon this Question.

If one needs to schedule a job interview for someone who wants to teach at a school. For the day of the interview, I need to schedule six different activities for the candidate:

• Meet with the Vice President of the school
• Talk with the Teachers in the department
• Tour the area/campus
• Teach a lesson
• take a 15 min break
• take a 30 min break

How many unique schedules can I create if I can't schedule a break at the beginning nor end? Assuming I must schedule each of the six activities at least once.

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Normally if the order didn't matter I would just say $6!$, but since the breaks can't be in the beginning or end, I'm at a bit of a loss.

Answer 1

Assuming the activities have to be scheduled exactly once each, you have six slots. Let's put the two breaks somewhere. How many choices are there, considering they can't be at the end or next to each other? There are few enough to count by hand. Now put the rest in order and use the available slots. Multiply and you are done.

Answer 2

Logically, the breaks must be at either

1. slots 2 and 4

or

2. slots 3 and 5

or

3. slots 2 and 5

There are 2 breaks to choose from and 4 activities to choose from, so

For case 1,

4 x 2 x 3 x 1 x 2 x 1 = 48

For case 2,

4 x 3 x 2 x 2 x 1 x 1 = 48

For case 3,

4 x 2 x 3 x 2 x 1 x 1 = 48

Case 1 + Case 2 + Case 3 = 144

Answer 3

Assuming each activity is supposed to be scheduled once only. number of ways to choose when both break will occur: C(4,2)x 2 number of ways for other activities to be arranged: 4! So final answer = 2xC(4,2)x4!=288