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Re: 2015 permutation X2 marathon Yep I got A winning in 1287 ways. Probably did something really stupid with the probability.

Re: 2015 permutation X2 marathon The problem is I can't find anything wrong with my method which is doing my head in. We have a bijection between arranging W's and L's and the ways Aristotle can win, but yet something isn't adding up.

Re: 2015 permutation X2 marathon Nevermind it's fine. I don't know anymore lol

Re: 2015 permutation X2 marathon I think I have to fix up my bijection actually.

Re: 2015 permutation X2 marathon Like I'm pretty confident it's 3/7 (if it isn't im going to look like the biggest dickhead)

Re: 2015 permutation X2 marathon Should we not be truncating after the 8th win though?

Re: 2015 permutation X2 marathon Arrange 8 W's and 5 L's for Aristotle \frac{13!}{8!~5!}=1287 Arrange 6 W's 7 L's for Socrates \frac{13!}{6!~7!}=1716 Answer is \frac{1287}{1287+1716} = \frac{3}{7}

Re: 2015 permutation X2 marathon I think the answer should be \frac{3}{7}

Re: 2015 permutation X2 marathon Like in reality the sequences WWWWWWWWWWWWW and WWWWWWWWLLLLL should be equivalent, but you are counting them as different by doing this.

Re: 2015 permutation X2 marathon You are though. Truncating after the 8th W gives a bijection between the ways Aristotle can win and the arrangements of 8 W's 5 L's.

Re: 2015 permutation X2 marathon Either I'm really derping with this question but surely the total outcomes isn't 2^{13}?

Re: 2015 permutation X2 marathon My way was: Aristotle needs to win 8 games, and can only lose at most 5. So take 8 W's and 5 L's and arrange them in a line to find the total ways Aristotle can win: \frac{13!}{8!~5!} This isn't a multiple of 595 though so...

Re: 2015 permutation X2 marathon Doing it my way gives a different answer? Might be doing something wrong subtly.

Alternatively, consider: \sum_{k=1}^{n} (a_kx+b_k)^2 (a_1^2+...+a_n^2)x^2+2(a_1b_1+...+a_nb_n)x+(b_1^2+...+b_n^2) This has discriminant \leq0 4(a_1b_1+...+a_nb_n)^2-4(a_1^2+...+a_n^2)(b_1^2+...+b_n^2) \leq 0 (a_1b_1+...+a_nb_n)^2 \leq (a_1^2+...+a_n^2)(b_1^2+...+b_n^2) Let a_k=\sqrt{x_k}...

For the second one, use the substitutions: a=\frac{x+z-y}{2} b=\frac{x+y-z}{2} c=\frac{y+z-x}{2} Then after simplifying the LHS we have: \frac{1}{2}(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y})-\frac{3}{2} But from the first inequality, we know that the thing in brackets above is greater...

Break it into: \frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+ \frac{b}{c} +\frac{c}{b} We know \frac{x}{y}+\frac{y}{x} \geq 2, so using this 3 times gets the result.

Re: HSC 2015 4U Marathon So why can you differentiate/integrate over the complex field?

Re: 4u trials CSSA That last question seems pretty easy?

HSC is similar unless they have changed how they mark since 2013.

Re: Semester 2 USYD Chatter Thread I just came out of my algebra lecture so I was probably thinking about something related to it lol. First week over for me, so far everything is actually quite fun. I haven't missed a lecture yet either so goooooo meeeeeeeeee. Although algebra is already...