Hi C,
I'll do what I can, but the explanation I made on page 128-129 is, to me, the most straightforward way to do it and I'm not sure how much I can improve on that but I'll try
The distribution here is one about the number of patients scheduled for each day. So, let's start with what we know:
- Number of patients = 9
How many days are they being scheduled for? 7
How many must be scheduled each day? At least 1
So, if you think about this as if you were in charge of scheduling. You've got 9 patients to assign over 7 days. How do you do it?
Well, the first thing is that you need 1 patient per day to satisfy the stated requirement. this would look like:
- Monday = 1
Tuesday = 1
Wednesday = 1
Thursday = 1
Friday = 1
Saturday = 1
Sunday = 1
That by itself assigns 7 of the 9 patients. So, what can we do with those other two patients? We can assign them anywhere, and we can assign them both to one day or we can split them up and assign them to different days. For simplicity sake, let's say we assigned both of the extra patients to one day, Friday:
- Monday = 1
Tuesday = 1
Wednesday = 1
Thursday = 1
Friday = 3
Saturday = 1
Sunday = 1
So, the question here is, can we ever get more than 3 patients on Friday (or any other day)? No, we've now assigned all 9 patients and there are no more left to assign. So, on Friday (or any day), the maximum number is 3. Thus, (C) is the correct answer in that problem.
Note that we could have assigned that 3 anywhere, to any day, but that every other day must then be 1 patient. That's where we get the unfixed 3-1-1-1-1-1-1 distribution from.
If the above doesn't make sense, try making up a schedule on your own. Create 9 patients and then assign them to different days using the "at least one" rule above. There are only two ways to do it!
Please let me know if that helps. Thanks!