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The answer to 2 is also wrong. It is just k! (n-k)!

The number of numbers without restriction is 5!/2! = 60. So the answer can't be 72. The correct calculation is 4! + 4!/2!.

Re: MX2 2015 Integration Marathon When doing that recurrence relation, it is a little bit slicker if you start with I_n + I_(n-2).

Re: MX2 2015 Integration Marathon Experimenting with latex - so this could be rubbish: - \sum_{k=1}^{9}\frac{\cot^{k}x}{k}

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Why don't you substitute to check? Also consider whether you have missed any solutions.

Re: HSC 2015 3U Marathon The question says "AT THE TOP OF ITS MOTION it touches the top vertex of the triangle."

Re: HSC 2015 3U Marathon As you said, it's a PARABOLA. The axis of symmetry is half way between the roots. That makes the triangle isosceles.

Re: HSC 2015 3U Marathon Since alpha MUST equal beta, why not write tan theta = 2 tan alpha ?

They are NOT the answers. Two answers are purely real. And all answers include a √2. The answers above don't work on substitution.

I don't see any need to use trig. Just write (z+1)²=2(z-1)² and (z+1)²=-2(z-1)². Then rearrange and solve each equation using the quadratic formula. What is more interesting is that the answers all lie on a circle centred at (3,0) with radius 2√2. I'm wondering if anyone can show why...

Re: 2015 permutation X2 marathon The probability they all go through one door is actually 1 / [n^(n-1)] Although you have my answer, I don't see how it follows from your first line.

Re: 2015 permutation X2 marathon See what happens with your answer when n=4.

That brings back good mammaries.

Ahh no - I have a reputation to protect.

Re: 2015 permutation X2 marathon The answer for anyone who wants to peek: 1 - n! / (n^n)

Does it apply to me?

Re: 2015 permutation X2 marathon Drsoccerball's solution is correct. 3!/2! = 3, and there most definitely is 3 ways to arrange 4,4,2 or 4,3,3. Another thought process (but same calculation): (For the 4,4,2 case): Pick the 2 first: 10C2 Pick the room for those two: 3 Pick one of the...

Re: 2015 permutation X2 marathon I can see your answer when I "reply with quote". It's hard to tell without proper formatting, but it seems to be correct.

Re: 2015 permutation X2 marathon * There are n balls in a bag, numbered with consecutive integers 1 to n. Show that the probability that if p balls are chosen that no chosen ball has number 1 to q is the same as the probability that if q balls are chosen that no chosen ball has number 1 to p.

From the board - "Pencil may be used only where specifically directed."