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\int x dx = \int x . \frac{d}{dx}(x) dx = x.x - \int x.\frac{d}{dx}(x) dx = x^2 - \frac{x^2}{2} = \frac{x^2}{2}

Because that scenario is exactly what addition of velocity vectors is all about.

I'm afraid this is not even a calculus question. It is just addition of vectors.

Actually, the form the equation was given in originally is general form.

Re: 2015 permutation X2 marathon Actually, it is not something HSC students need to know. I asked the question on the basis that this result would have been derived in an earlier part.

Re: 2015 permutation X2 marathon Let T_n be the number of ORDERED subsets of n distinct objects. Show that as n approaches infinity, T_n approaches e.n!

Re: 2015 permutation X2 marathon Yep, that's it. One of my students came up with that a couple of years ago, but I couldn't remember it.

Re: 2015 permutation X2 marathon Give a combinatorics reason (NOT a binomial theorem reason) why n.2^(n-1) = nC1 + 2.nC2 + 3.nC3 + ... + n.nCn [I actually don't recall the reason and am trying to remind myself. So a prompt would be useful rather than the entire answer.]

Re: 2015 permutation X2 marathon (6×5)×(5×4)×(4×3) Edit: Yes - forgot the multiple counting.

Re: 2015 permutation X2 marathon Nup. 1 - 364/365 . 363/365 . 362/365 . ... . 336/365

My wording does not precisely reflect what I was trying to say. It should be: The points of intersection of a function and its inverse don't have to lie on y=x or y=-x.

Re: 2015 permutation X2 marathon Well for starters, that wasn't the actual question. Can you answer the question with the Friday info included?

Re: 2015 permutation X2 marathon Well - yes - and that probably helps with the latter part of the question. But here its just a matter of listing out the cases: GG, GB, BG, BB and then eliminating the BB case.

Re: 2015 permutation X2 marathon If anyone is having difficulty, you would probably have a better chance of seeing the concept if the question was rephrased in terms of counting. Focusing on the case without the Friday info: "Of all the two child families with at least one girl, what fraction...

Re: 2015 permutation X2 marathon No it's not. We are not told that "she" is born on a Friday, we are told that one of them is born on a Friday. Being told that one is a "girl Friday" is not enough to distinguish between the children, because the other could be also be a girl born on a...

Re: 2015 permutation X2 marathon It would be 1/2 if you were told that a particular child (say the first-born) was a girl born on a Friday. It is not 1/2 because you are not told which child - only that one of them satisfies this condition. Not sure we needed the Friday part though - it would...

Re: 2015 permutation X2 marathon No, I'm saying that if the colour of the first ball had been noted, then the answer would be different. It wasn't noted, so he is correct.

Re: 2015 permutation X2 marathon It's kind of like quantum mechanics - when you look at the first marble you change the answer!

What do you mean they're not out yet? You do know that not all schools have the same trial fortnight, right? Some schools had results out 2 weeks ago.

Re: 2015 permutation X2 marathon Warning: beyond high school difficulty 12 counters numbered 1 to 12 are to be distributed one at a time in ascending numerical order between two distinguishable boxes A and B so that we finish with 6 in each box. In how many ways can this be done so that at...