Drsoccerball
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HSC 2016 MX2 Combinatorics Marathon (archive) (1 Viewer)
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Paradoxica
-insert title here-
Drsoccerball
Well-Known Member
- Joined
- May 28, 2014
- Messages
- 3,657
- Gender
- Undisclosed
- HSC
- 2015
Paradoxica
-insert title here-
Re: HSC 2016 MX2 Combinatorics Marathon
27 unit-cubes are assembled into a single 3-by-3-by-3 cube.
The surface of this larger cube is painted red.
The cube falls apart, and a blind man reassembles it.
What is the probability that the reassembled cube again has its entire surface painted red?
27 unit-cubes are assembled into a single 3-by-3-by-3 cube.
The surface of this larger cube is painted red.
The cube falls apart, and a blind man reassembles it.
What is the probability that the reassembled cube again has its entire surface painted red?
KingOfActing
lukewarm mess
Re: HSC 2016 MX2 Combinatorics Marathon
? If that's right I'll edit in my though process 
Is the answer
Re: HSC 2016 MX2 Combinatorics Marathon
But I have worked out an answer that is about 10 to the power of 20 smaller than yours!
I believe those factorials you have should actually be in the numerator. And don't forget you have to take into account the orientation of each cube - just because a cube is in the right position doesn't mean its red faces are facing outwards.
I'll give you my simplified answer (so it doesn't give an indication of my logic):
(6!8!12!)/(27! times 24^18)
It is the 24^18 which has been simplified.
(Not 100% sure it is right, but pretty confident yours isn't)
I found this question on an online forum and they had no definitive answer.Is the answer
But I have worked out an answer that is about 10 to the power of 20 smaller than yours!
I believe those factorials you have should actually be in the numerator. And don't forget you have to take into account the orientation of each cube - just because a cube is in the right position doesn't mean its red faces are facing outwards.
I'll give you my simplified answer (so it doesn't give an indication of my logic):
(6!8!12!)/(27! times 24^18)
It is the 24^18 which has been simplified.
(Not 100% sure it is right, but pretty confident yours isn't)
Paradoxica
-insert title here-
Re: HSC 2016 MX2 Combinatorics Marathon
If they couldn't arrive at a consensus what makes you think we will?
I personally want to find the experimental probability but I don't have the programming knowledge for it. It's been a while.
Care to state the forum?
If they couldn't arrive at a consensus what makes you think we will?
I personally want to find the experimental probability but I don't have the programming knowledge for it. It's been a while.
Re: HSC 2016 MX2 Combinatorics Marathon
But if my answer is correct (and I am reasonably confident it is) then you would need to run a simulation about 10^37 times just to get one success.
How long are you planning on living?
I'm afraid I can't remember exactly where I found it.
But if my answer is correct (and I am reasonably confident it is) then you would need to run a simulation about 10^37 times just to get one success.
How long are you planning on living?
Paradoxica
-insert title here-
Re: HSC 2016 MX2 Combinatorics Marathon
So will the end of time cause your death, or will your death cause time to stop?
Paradoxica
-insert title here-
KingOfActing
lukewarm mess
Re: HSC 2016 MX2 Combinatorics Marathon
The way I got that was
- i.e. choose 8 corners, 12 sides and 6 centers but I feel like I'm missing lots
The way I got that was
Paradoxica
-insert title here-
Re: HSC 2016 MX2 Combinatorics Marathon
Does that account for multiplicity?The way I got that was
Re: HSC 2016 MX2 Combinatorics Marathon
You would need to account for making sure the cubelets are rotated the correct way too (so that their painted faces are outside).The way I got that was
Re: HSC 2016 MX2 Combinatorics Marathon
Any cube can be oriented in 24 ways.
Only 3 of those ways will cause a corner cube to have its red faces on the outside.
2 of those ways will cause an edge cube to have its red faces on the outside.
4 of those ways will cause a face-centre cube to have its red faces on the outside.
As I said, you're not accounting for the different orientations of the cubes.The way I got that was
Any cube can be oriented in 24 ways.
Only 3 of those ways will cause a corner cube to have its red faces on the outside.
2 of those ways will cause an edge cube to have its red faces on the outside.
4 of those ways will cause a face-centre cube to have its red faces on the outside.
Paradoxica
-insert title here-
Re: HSC 2016 MX2 Combinatorics Marathon
Consider an alphabet with n different available letters.
Let P(k) be the number of ways you can use exactly k different letters in an n letter word.
i) Explain why
 \binom{n}{k} = n^n)
ii) Show that
Let the Score of a word, X, be defined as 1/(1+#(X)), where #(X) is the number of letters that were not used by the word X.
iii) Show that the sum of all the Scores, S, over all possible n letter words, is given by:
}{n+1-k} \binom{n}{k})
iv) Hence, evaluate S in closed form.
Consider an alphabet with n different available letters.
Let P(k) be the number of ways you can use exactly k different letters in an n letter word.
i) Explain why
ii) Show that
Let the Score of a word, X, be defined as 1/(1+#(X)), where #(X) is the number of letters that were not used by the word X.
iii) Show that the sum of all the Scores, S, over all possible n letter words, is given by:
iv) Hence, evaluate S in closed form.
Paradoxica
-insert title here-
Re: HSC 2016 MX2 Combinatorics Marathon
How to summon Carrotsticks???
@Carrotsticks
idk
How to summon Carrotsticks???
@Carrotsticks
idk
KingOfActing
lukewarm mess
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