Terry Lee's Solutions 2015 Extension 2 HSC (1 Viewer)

Carrotsticks

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Haha woops. My inequality for the x^2 and x^2+1 is the wrong way around, woops!

But the idea is still there of course.

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tondog

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But logically it does make sense, as n tends to infinity, the number of possible squares for the black counters to be in the same column increases.
I'm not too sure either but I think its also possible to show that algebraically.
Pn=(n^q)(nq-q)!(q!)/(nq)!
= (n^q)(q!)/(nq)(nq-1).....(nq-q+1)

=(n n n n n.......)(q!)/(nq)(nq-1).....(nq-q+1)

=(nq)(nq-n)(nq-2n)......)/(nq)(nq-1).....(nq-q+1)

Now there is q terms on top and (nq)(nq-1).......(nq-q+1) must have more than q terms

tf it is = to roughly 1/(nq-1)(nq-2).....)
q>0 and as n tend to infinity Pn=0
Sorry I dont know how to use w/e the program is, I only came here out of interest because of this question as well.
Correct me if I'm wrong or if I missed something, thanks.
 
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RealiseNothing

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But logically it does make sense, as n tends to infinity, the number of possible squares for the black counters to be in the same column increases.
I'm not too sure either but I think its also possible to show that algebraically.
Pn=(n^q)(nq-q)!(q!)/(nq)!
= (n^q)(q!)/(nq)(nq-1).....(nq-q+1)!

=(n n n n n.......)(q!)/(nq)(nq-1).....(nq-q+1)

=(nq)(nq-n)(nq-2n)......)/(nq)(nq-1).....(nq-q+1)

Now there is q terms on top and (nq)(nq-1).......(nq-q+1) must have more than q terms

tf it is = to roughly 1/(nq-1)(nq-2).....)
q>0 and as n tend to infinity Pn=0
Sorry I dont know how to use w/e the program is, I only came here out of interest because of this question as well.
Correct me if I'm wrong or if I missed something, thanks.
The answer to the question is

Since there are infinite spaces in each column, you can treat each column as a number (say 1, 2, ... , q).

The question is equivalent to "if I have a bag with the numbers 1 up to q in it, and I pick out q numbers with replacement, what is the probability that I pick out each number exactly once?"
 

glittergal96

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The answer to the question is

Since there are infinite spaces in each column, you can treat each column as a number (say 1, 2, ... , q).

The question is equivalent to "if I have a bag with the numbers 1 up to q in it, and I pick out q numbers with replacement, what is the probability that I pick out each number exactly once?"
Would be wary about writing this solution in an MX2 exam.

Finding the limit of what the probability is as n gets big is not necessarily the same as just treating the problem as taking place on an infinity x q grid. Furthermore what does it even mean to choose a square randomly from an infinite grid of this form, and why is it equally likely you will be in each column?

You might have some intuition for these things, and your intuition may lead to the correct answer, but it would be difficult to properly justify this approach using MX2 language and methods.

IMO a squeeze theorem approach would be far safer.
 

RealiseNothing

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But logically it does make sense, as n tends to infinity, the number of possible squares for the black counters to be in the same column increases.
I'm not too sure either but I think its also possible to show that algebraically.
Pn=(n^q)(nq-q)!(q!)/(nq)!
= (n^q)(q!)/(nq)(nq-1).....(nq-q+1)!

=(n n n n n.......)(q!)/(nq)(nq-1).....(nq-q+1)

=(nq)(nq-n)(nq-2n)......)/(nq)(nq-1).....(nq-q+1)

Now there is q terms on top and (nq)(nq-1).......(nq-q+1) must have more than q terms

tf it is = to roughly 1/(nq-1)(nq-2).....)
q>0 and as n tend to infinity Pn=0
Sorry I dont know how to use w/e the program is, I only came here out of interest because of this question as well.
Correct me if I'm wrong or if I missed something, thanks.
The problem with that logic is ALL the columns now have infinitely many squares, not just one. Yes there are an infinite amount of squares in one column that the black counter can be put in, but there is also a infinite amount of squares NOT in that column. The logic breaks down when you are dealing with the competing infinities (not really sure what else to call it).

In terms of your algebraic proof, I don't understand how the bolded is true? There are q terms on the bottom too.

I also don't see how you got to the line "it is roughly = to ...", why did you just cancel all the top terms out?
 

RealiseNothing

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Would be wary about writing this solution in an MX2 exam.

Finding the limit of what the probability is as n gets big is not necessarily the same as just treating the problem as taking place on an infinity x q grid. Furthermore what does it even mean to choose a square randomly from an infinite grid of this form, and why is it equally likely you will be in each column?

You might have some intuition for these things, and your intuition may lead to the correct answer, but it would be difficult to properly justify this approach using MX2 language and methods.

IMO a squeeze theorem approach would be far safer.
I would have thought that was the approach they were looking for though. It might not be completely rigorous but it is the HSC and they aren't always frantic about those things.

Curious to see if they would accept this solution.
 

glittergal96

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I would have thought that was the approach they were looking for though. It might not be completely rigorous but it is the HSC and they aren't always frantic about those things.

Curious to see if they would accept this solution.
I guess we will have to wait till the exam notes come out from BOS to know for sure, but usually the questions at this point in the paper are written by people who know well enough that the answers they are designed to elicit are actually quite rigorous.

(At least rigorous as deductions from the theory learned in MX2. Looking back at final questions from previous exams involving analysis, things like the squeeze law are very commonly used to draw the final conclusion about the limit. This replacement of a finite combinatorial situation by a vaguely specified infinite one in the limit would be pretty out of character I think.)

The infinite grid gives you a pretty easy and good reason to expect the answer to come out as it does, but I would be very hesitant to use it as actual justification for the answer.
 
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tondog

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Sorry I just realised (nq)(nq-1).....(nq-q+1) has q terms
Ok i just realised how to solve it algebraically

n^q(q!)/(nq)(nq-1)(nq-2).....(nq-q+1)

(nq)(nq-1)(nq-2).....(nq-q+1)=n^q(q)(q-1/n)(q-2/n).......

tf (q!)/(q)(q-1/n)(q-2/n).......
n tends to infinity
=q!/q^q
 

glittergal96

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Sorry I just realised (nq)(nq-1).....(nq-q+1) has q terms
Ok i just realised how to solve it algebraically

n^q(q!)/(nq)(nq-1)(nq-2).....(nq-q+1)

(nq)(nq-1)(nq-2).....(nq-q+1)=n^q(q)(q-1/n)(q-2/n).......

tf (q!)/(q)(q-1/n)(q-2/n).......
n tends to infinity
=q!/q^q
Yep, this is good and valid. :)
 

InteGrand

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I would have thought that was the approach they were looking for though. It might not be completely rigorous but it is the HSC and they aren't always frantic about those things.

Curious to see if they would accept this solution.
Knowing the HSC, I think they just wanted people to answer the Q. by finding the limit of a ratio of same degree polynomials. That's probably why they had the part (ii) (and gave you the expression for it, so you could still do the next part if you didn't manage to get (ii)). Then they just wanted or expected you to find the limit of this expression.
 

glittergal96

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Is it ok to simplify it further and say (q-1)! / q^(q-1) ?
Yeah, sure. There's no way they would mark that wrong lol.

As for "simplification", that is debatable,as it requires more writing rather than less. This fraction will usually be far from simplest form anyway.
 

Ekman

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Yeah, sure. There's no way they would mark that wrong lol.

As for "simplification", that is debatable,as it requires more writing rather than less. This fraction will usually be far from simplest form anyway.
Under exam conditions I initially thought I was wrong, so I tried simplifying it as much as possible and eventually just ended up with saying (q-1)! / q^(q-1) as my final result, and so seeing that most solutions just kept it as q! / q^q, I got worried that my 'simplified' form may not be accepted lol.
 

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