Past Paper Questions for 3u (1 Viewer)

frankfurt

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Hi,

Could you please help me out with these past paper questions?

1602326756562.png
The answer is C but the answer i get is 362304

1602326900122.png
There was actually no answer for this one in the answer key for this one. i think they might've skipped it

thanks in advance:))
 

devtrivedi

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Yeh the answer is C
To find the arrangement, do the total number of ways (9!) - the number of arrangements where no boys are together (4! x 5!)
 

devtrivedi

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and for Part (e) f(g(x))=x for every function where g(x) is the inverse function of f(x)
 

YonOra

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Why isn't it 2! (the two different ways you can seat the students) x 8! (the remaining seats and how you can seat them). Admittedly, i've half forgotten this stuff and have only done Yr11 content with it, and its a bit sus that im' not using the (n-1)! formula. Can anyone explain?
 

Pikapizza

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Why isn't it 2! (the two different ways you can seat the students) x 8! (the remaining seats and how you can seat them). Admittedly, i've half forgotten this stuff and have only done Yr11 content with it, and its a bit sus that im' not using the (n-1)! formula. Can anyone explain?
The question states how many ways you can do if at least two students sit together so either you take cases (if 2 students sit together, if 3 sit together, if 4..., if 5... and so on)
OR you do as @devtrivedi did which is much quicker, you do: total - arrangements of the boys all separated (do insertion method like above)

Edit: there would be many more ways to do these questions but when you see an “at least” it is most common for it to be either cases or total-unfavourable method
 

CM_Tutor

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Why isn't it 2! (the two different ways you can seat the students) x 8! (the remaining seats and how you can seat them). Admittedly, i've half forgotten this stuff and have only done Yr11 content with it, and its a bit sus that im' not using the (n-1)! formula. Can anyone explain?
Had the question been to seat the 10 people around a circular table with two specific people seated together, the solution would match your suggestion:
  • Seat the first of the two specific people in 1 way (as all positions at an empty circular table are identical under rotation)
  • Seat the second of the two specific people in 2 ways (in one of the two seats adjacent to the first specific person)
  • There are 8 people remaining and 8 seats, so there are 8! arrangements.
  • Thus, total number of arrangements is 2 x 8! = 80,640
However, the question has two groups of 5 people (students and parents) and we need at least two students together. We could do this by counting up the cases:
  1. with exactly two students together (and the other three students not together with the first two, but possibly together with each other)
  2. then, with exactly three students together (and the other two together or separate but not together with the first three)
  3. then, with four students together
  4. then, with all five students together
but that is long and involved.

We are better off looking to the complementary event, which is the single case with no students together. This can only be done if the seating alternates parent - student - parent - student etc.
  • Total number of arrangements of 10 people at a circular table is 9! = 362,880
  • Seating alternately, we seat a student first in 1 way, then the other four students have only four possible seats, so seat them in 4! ways. The remaining five seats go to the five parents in 5! ways
  • So, seating alternatively can be done in 1 x 4! x 5! = 2,880 ways
  • So, at least two students together has 9! - 4! x 5! = 360,000
Intuitively, the likelihood of there being at least one pair of students together if seated randomly is quite high as the only way that it doesn't happen is if the result is an alternate seating pattern - and alternate seating is unlikely. Two students together is also less likely if it is a specific pair who must be together. Looking at the above answers, we have:



This fits the pattern that is expected, and shows that the specific two together interpretation that you are thinking about has a probability of around 22% compared to over 99% for at least two of the students, not specifying which, are together.
 

CM_Tutor

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The question states how many ways you can do if at least two students sit together so either you take cases (if 2 students sit together, if 3 sit together, if 4..., if 5... and so on)
OR you do as @devtrivedi did which is much quicker, you do: total - arrangements of the boys all separated (do insertion method like above)

Edit: there would be many more ways to do these questions but when you see an “at least” it is most common for it to be either cases or total-unfavourable method
Pikapizza is correct that there is a difference between exactly and at least two students. However, the approach I addressed above leading to 2 x 8! requires the two students in question to be specified. The number of arrangements with the two together being any pair of students would be different again.
 

YonOra

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Had the question been to seat the 10 people around a circular table with two specific people seated together, the solution would match your suggestion:
  • Seat the first of the two specific people in 1 way (as all positions at an empty circular table are identical under rotation)
  • Seat the second of the two specific people in 2 ways (in one of the two seats adjacent to the first specific person)
  • There are 8 people remaining and 8 seats, so there are 8! arrangements.
  • Thus, total number of arrangements is 2 x 8! = 80,640
However, the question has two groups of 5 people (students and parents) and we need at least two students together. We could do this by counting up the cases:
  1. with exactly two students together (and the other three students not together with the first two, but possibly together with each other)
  2. then, with exactly three students together (and the other two together or separate but not together with the first three)
  3. then, with four students together
  4. then, with all five students together
but that is long and involved.

We are better off looking to the complementary event, which is the single case with no students together. This can only be done if the seating alternates parent - student - parent - student etc.
  • Total number of arrangements of 10 people at a circular table is 9! = 362,880
  • Seating alternately, we seat a student first in 1 way, then the other four students have only four possible seats, so seat them in 4! ways. The remaining five seats go to the five parents in 5! ways
  • So, seating alternatively can be done in 1 x 4! x 5! = 2,880 ways
  • So, at least two students together has 9! - 4! x 5! = 360,000
Intuitively, the likelihood of there being at least one pair of students together if seated randomly is quite high as the only way that it doesn't happen is if the result is an alternate seating pattern - and alternate seating is unlikely. Two students together is also less likely if it is a specific pair who must be together. Looking at the above answers, we have:



This fits the pattern that is expected, and shows that the specific two together interpretation that you are thinking about has a probability of around 22% compared to over 99% for at least two of the students, not specifying which, are together.
Thank you, this was really helpful!
 

CM_Tutor

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so then is the answer just 3?
No.

Since is the inverse of , so long as is within the appropriate domain. It follows, so long as is in the domain, that , but we seek to find .

First, checking the domain, in this case we have



and so increases throughout its domain, which is - and so can be inverted for all .

The question directs that you consider the derivative of . Since is within the domain of inversion, we know that . Our goal, remember, is the value of :



So, we need the value of , and given , we need the value of .





 

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