awesome245
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- 2023
nvm I think I'm wrong
Start from a teacher.How is it possible for only two students to sit next to each other, as there is no combination that works
I might be reading the question wrong, but doesn't this not work as a maximum of two students can be sat next to each other (not two pairs)?Starting from a teacher, the arrangement must be T-S-S-T-S-S-T-S-and we are back to the original teacher, or two variations T-S-S-T-S-T-S-S- and T-S-T-S-S-T-S-S-.
Seat first teacher in 1 way
Choose one of these 3 arrangements
Seat the two remaining teachers in 2 ways
Seat the students in 5! ways
Arrangements = 1 x 3 x 2 x 5! = 3! x 5!
To seat the students as one pair and three singles means many more possible arrangements, as it needs to be SS-X-S-X-S-X-S-X-, where X is not a student.Oh I thought it was a maximum of two students total, not any number of pairs of two students, thanks
lmao yeah I think it was badly wordedTo seat the students as one pair and three singles means many more possible arrangements, as it needs to be SS-X-S-X-S-X-S-X-, where X is not a student.
Available to fill the four non-student positions are three teachers, so at least one of the teachers must be sub-divided. Now, a single teacher could be halved but halving vertically (into a left and right half) would be different from halving horizontally into a top and bottom half, two consider just two directions. Then the spacing objects need not be equally sized, so we could use just a fraction of a teacher (say, a leg) for the fourth position... but is a left leg identical to a right leg, or are these yet more possibilities needing to be accounted for? And, do we have to triple the options as deconstructing one teacher is different from another? Perhaps one of the students will volunteer a school bag as a separator and spare the teachers from subdivision... but then, if that student is adjacent to their own bag, does it return to being a single unit, and so the bag only acts as a separator when not adjacent to its owner?
The possibilities under this interpretation are endless... I think the wording needs to be considered as restricting student groupings to a maximum size of two without limiting the number of such groupings.
this question also relates to cross product too. you can only do cross product in R3 but R3 contains R2 so just do a = (a1,a2,0), b=(b1,b2,0) --> Area of parallelogram spanned by a and b = |a x b| = |a1b2 - a2b1| --> triangle area is 1/2 thatView attachment 40843
Ever heard of the shoelace formula?
Some people viewing this would instantly recognise a1b2-a2b1 as being a determinant.
The result in this paper can be generalised to find areas of polygons, not just triangles:
https://en.wikipedia.org/wiki/Shoelace_formula
It can be done in 7 dimensions too: https://en.wikipedia.org/wiki/Seven-dimensional_cross_productyou can only do cross product in R3
imo 2023do you guys think 2022 or 2023 was harder? I'm praying on this year's scaling. what do you think is the e4 raw mar cut off?
The question also said "or otherwise"Area of parallelogram spanned by a and b = |a x b| = |a1b2 - a2b1| --> triangle area is 1/2 that
just checked solutions, should hopefully be 66/70 unless I wrote something wrong and forgot.carrotsss how did u go? 70/70?????/
woooooo noice!!!!just checked solutions, should hopefully be 67/70 unless I wrote something wrong and forgot.
nah he just smartThat must’ve taken some commitment. Nice.
lmao dont snitch on urself then xDnah he just smart
@TankKuno i managed to lose a mark on that projectile collision question due to silly error mr mai is gonna be so mad bruh we did like 5000 of those questions in class
So if you wrote it out in full it would beThe question also said "or otherwise"
You just gave an "otherwise" method. Don't know if they would award full marks for it. They should though.
The otherwise method is much easier too.