Mathematics Extension 1 Predictions/Thoughts (1 Viewer)

[awesome245]

nvm I think I'm wrong

 

[awesome245]

How is it possible for only two students to sit next to each other, as there is no combination that works

 

[Luukas.2]

How is it possible for only two students to sit next to each other, as there is no combination that works

Start from a teacher.

The arrangement must be T-SS-T-SS-T-S-(and back to the original teacher), or one of two variations: T-SS-T-S-T-SS- and T-S-T-SS-T-SS-.

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!

 

[awesome245]

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!

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)?

 

[awesome245]

Oh I thought it was a maximum of two students total, not any number of pairs of two students, thanks

 

[tywebb]

.

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

"Shoelace formula" is really a misnomer. What it actually is is an algorithm for very quickly calculating the determinants - without using the formula.

 

[Luukas.2]

Oh I thought it was a maximum of two students total, not any number of pairs of two students, thanks

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.

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, considering just two possible planes of sub-division. 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.

 

[awesome245]

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.

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.

lmao yeah I think it was badly worded

 

[notme123]

View 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

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 that

 

[blob063540]

yeh i think lucid's working out for Q10 is wrong, bc u cant just say 2! for organising the teachers bc they are aligned in an isosceles triangle formation, and thus where you place the first teacher matters. What I did instead was chose a student (5C1) to place down as the student who was on their own and then did 3! for the teachers.

Also, he grouped the students with 3!, which i believe is not a valid way of counting the number of ways to group the students, as u cant just distribute 5 students in 3 spots as if ur distributing 3 students in 3 spots, and then just multiply by 5!

I instead, having already placed down a student chose 2 students to place in one of the sections to place a pair (4C2) and then distributed them within that allocation (2!). Following this, the remaining students were then distributed within the other section to place the pair (2!)

Overall I got B instead of C

My explanation probs sounds hella confusing but I just want someone else to confirm that B is correct

 

[switchpen]

do 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?

 

[tywebb]

you can only do cross product in R3

It can be done in 7 dimensions too: https://en.wikipedia.org/wiki/Seven-dimensional_cross_product

 

[carrotsss]

do 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?

imo 2023

 

[TankKuno]

carrotsss how did u go? 70/70?????/

 

[tywebb]

Area of parallelogram spanned by a and b = |a x b| = |a1b2 - a2b1| --> triangle area is 1/2 that

The 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.

 

[carrotsss]

carrotsss how did u go? 70/70?????/

just checked solutions, should hopefully be 66/70 unless I wrote something wrong and forgot.

nvm 66 just found a mistake

 

[TankKuno]

just checked solutions, should hopefully be 67/70 unless I wrote something wrong and forgot.

woooooo noice!!!!

 

[synthesisFR]

That must’ve taken some commitment. Nice.

nah 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

 

[TankKuno]

nah 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

lmao dont snitch on urself then xD

 

[tywebb]

The 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.

So if you wrote it out in full it would be