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Actually he probably did. I didn't read the post properly and missed "of the form".

Integration and Volumes are the easiest topics in Ext 2.

Actually the Ext 2 syllabus specifically states that these questions are NOT part of the course: "Recurrence relations such as ....... which involve more than one integer parameter, are excluded."

I guess there were 1200 farewell handshakes. (Followed by 20C2 + 60C2 high fives)

Oh really. I didn't know that. Even after almost 30 years of teaching maths I still learn new things every day.

Actually, this question is not clear. If the 1.14% is an average growth rate then the calculation given by jamesh1234 is correct. If it is an instantaneous growth rate then the calculation is 23246682 * e^(0.0114)

I don't think BOSTES cares. They also don't care about giving teachers time to learn long forgotten material.

Some time after we actually see a syllabus. And I'm pretty sure there won't be a new Coroneos.

If they are relevant, why not?

You *COULD*, but precisely who is hurt by posting here? I don't understand why it is a problem for anyone.

But this guy *DID* show an interest. And so did someone else.

Why not?

I hate the name "Perms and Combs" for this topic, for two reasons: (1) It suggest that all counting is done using just these two techniques (2) It sounds like a hairdressing course

Haven't you answered your own question ... do past papers. All you need is the solutions to help you when you can't answer them yourself.

Re: HSC 2016 MX2 Combinatorics Marathon Hmm ... not sure how to interpret that comment.

Re: HSC 2016 MX2 Combinatorics Marathon Well I am the teacher here. I have to limit my showing off for the benefit of the students .... difficult as that might be.

Re: HSC 2016 MX2 Combinatorics Marathon Just checking - is the answer 1/4 ? (If it is, I'll leave it for someone else to provide a solution)

Re: HSC 2016 MX2 Combinatorics Marathon Yep, that's the one.

Re: HSC 2016 MX2 Combinatorics Marathon It hadn't occurred to me that my question was ambiguous. My intention was that the original interval remains length 1, and the length of the middle segment would change to get a probability of 1/3.

Re: HSC 2016 MX2 Combinatorics Marathon Open to all. And the method IS related to a past HSC question. \text{It is desired to cut the interval }0 \leqslant x \leqslant 1 \text{ into three segments.} \text{Two numbers }a \text{ and }b \text{ are chosen randomly, satisfying }0 \leqslant a,b...