HSC 2015 MX2 Permutations & Combinations Marathon (archive) (1 Viewer)

[Drsoccerball]

Re: 2015 permutation X2 marathon

[*]

Including the empty set, how many subsets of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} do not contain any consecutive integers?

Answer: 377


(Since I am not going to count what question we are up to, perhaps we should mark new questions as I have - in the largest font)

I keep getting 298 just to make sure is the qustion asking how many numbers can be made such that none are consecutive and that includes one digit numebrs 2 digits 3 digits etc...?
And consecutive meaning right after one another or if my first number 2 can i pick 4?

 

[braintic]

Re: 2015 permutation X2 marathon

I keep getting 298 just to make sure is the qustion asking how many numbers can be made such that none are consecutive and that includes one digit numebrs 2 digits 3 digits etc...?
And consecutive meaning right after one another or if my first number 2 can i pick 4?

You can have any set of just one number.
You can't have {2,3}, {2,3,4} or {2,4,6,8,10,11}.
You can have {2,4}, {2,4,6,8,10,12}, {1, 9, 12}.

 

[rand_althor]

Re: 2015 permutation X2 marathon

[*]

Including the empty set, how many subsets of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} do not contain any consecutive integers?

Answer: 377

The set {} has 1 subset with no consecutive integers including the empty set. The set {1} has 2, the set {1,2} has 3, the set {1,2,3} has 5. The amount of subsets with no consecutive integers increases as the amount of numbers in the set increases, in accordance with the Fibonacci sequence such that when there are n terms in a set, there are F_n+2 subsets with no consecutive integers (where F_n is the nth term in the Fibonacci sequence). So for the set {1,2,3,4,5,6,7,8,9,10,11,12}, there are F₁₄ = 377 subsets with no consecutive integers.

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

The set {} has 1 subset with no consecutive integers including the empty set. The set {1} has 2, the set {1,2} has 3, the set {1,2,3} has 5. The amount of subsets with no consecutive integers increases as the amount of numbers in the set increases in accordance with the Fibonacci sequence, such that when there are n terms in a set, there are F_n+2 subsets with no consecutive integers (where F_n is the nth term in the Fibonacci sequence). So for the set {1,2,3,4,5,6,7,8,9,10,11,12}, there are F₁₄ = 377 subsets with no consecutive integers.

I stil dont get what the questions asking...

 

[braintic]

Re: 2015 permutation X2 marathon

I stil dont get what the questions asking...

A subset is any group of any size selected from the original set. That includes the empty set (size zero) and the entire original set.

You should already be able to calculate the total number of subsets that can be formed from a set of 12 objects.

But we are only looking for subsets that don't include any pair of consecutive integers.

 

[rand_althor]

Re: 2015 permutation X2 marathon

I stil dont get what the questions asking...

Okay consider if the question was instead asking for the set {1,2,3,4}. A subset from this set is a set which consists of the elements of the original set. So the subsets which have do not have any consecutive integers from this set are:
With 0 elements: {} - the empty set.
With 1 elements: {1}, {2}, {3}, {4} - these are the only four possibilities.
With 2 elements: {1,3}, {1,4}, {2,4} - here we cannot have {1,2} or {2,3} etc. as the integers in the subset are consecutive. Also, {1,3} and {3,1} are the same subset so we only include one of them.
With 3 elements: There are no possibilities here. e.g. {1,3,4} does not work as 3 and 4 are consecutive.
With 4 elements: The only case is {1,2,3,4}, but this is not valid as the integers are consecutive.
In total for this set, there are 8 possibilities.

Does that help?

 

[braintic]

Re: 2015 permutation X2 marathon

The set {} has 1 subset with no consecutive integers including the empty set. The set {1} has 2, the set {1,2} has 3, the set {1,2,3} has 5. The amount of subsets with no consecutive integers increases as the amount of numbers in the set increases, in accordance with the Fibonacci sequence such that when there are n terms in a set, there are F_n+2 subsets with no consecutive integers (where F_n is the nth term in the Fibonacci sequence). So for the set {1,2,3,4,5,6,7,8,9,10,11,12}, there are F₁₄ = 377 subsets with no consecutive integers.

Yes, but you need to explain why they follow the Fibonacci sequence.
Seeing that the pattern exists is only the first step.

 

[InteGrand]

Re: 2015 permutation X2 marathon

 

[braintic]

Re: 2015 permutation X2 marathon

[*]

Aristotle and Socrates play in the final of the World Championship of Thong Throwing.
They each throw a thong, and the person who throws the furthest wins one game.
The winner of the championship is the first to win 10 games.
Currently, Socrates leads 4 games to 2.
If each player has an equal chance of winning a particular game, find the probability that Aristotle wins.

Answer: 595/2048

[Please do using counting techniques, not 2U probability]

 

[Ekman]

Re: 2015 permutation X2 marathon

[*]

Aristotle and Socrates play in the final of the World Championship of Thong Throwing.
They each throw a thong, and the person who throws the furthest wins one game.
The winner of the championship is the first to win 10 games.
Currently, Socrates leads 4 games to 2.
If each player has an equal chance of winning a particular game, find the probability that Aristotle wins.

Answer: 595/2048

[Please do using counting techniques, not 2U probability]

What are 'counting techniques', because I solved this question through considering cases.

 

[faisalabdul16]

Re: 2015 permutation X2 marathon

[*]

Aristotle and Socrates play in the final of the World Championship of Thong Throwing.
They each throw a thong, and the person who throws the furthest wins one game.
The winner of the championship is the first to win 10 games.
Currently, Socrates leads 4 games to 2.
If each player has an equal chance of winning a particular game, find the probability that Aristotle wins.

Answer: 595/2048

[Please do using counting techniques, not 2U probability]

Taking the extreme case where Socrates wins 5 consecutive games (putting him in the lead of 9 games to 2), this means that for Aristotle to win, he must win the next 8 games. What does this mean? That Aristotle will win in either 13 or less games.

The total number of outcomes is <a href="http://www.codecogs.com/eqnedit.php?latex=2^{13}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?2^{13}" title="2^{13}" /></a>

Now - Aristotle can afford to lose 1, 2, 3, 4 or 5 games and still win.

<a href="http://www.codecogs.com/eqnedit.php?latex={13\choose1}&space;&plus;&space;{13\choose2}&space;&plus;&space;{13\choose3}&space;&plus;&space;{13\choose4}&plus;&space;{13\choose5}&space;&plus;&space;1&space;=&space;2380&space;\\&space;\\&space;P(Aristotlte\&space;wins)&space;=&space;\frac{2380}{8192}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?{13\choose1}&space;&plus;&space;{13\choose2}&space;&plus;&space;{13\choose3}&space;&plus;&space;{13\choose4}&plus;&space;{13\choose5}&space;&plus;&space;1&space;=&space;2380&space;\\&space;\\&space;P(Aristotlte\&space;wins)&space;=&space;\frac{2380}{8192}" title="{13\choose1} + {13\choose2} + {13\choose3} + {13\choose4}+ {13\choose5} + 1 = 2380 \\ \\ P(Aristotlte\ wins) = \frac{2380}{8192}" /></a>

= 595/2048

 

[braintic]

Re: 2015 permutation X2 marathon

Taking the extreme case where Socrates wins 5 consecutive games (putting him in the lead of 9 games to 2), this means that for Aristotle to win, he must win the next 8 games. What does this mean? That Aristotle will win in either 13 or less games.

The total number of outcomes is <a href="http://www.codecogs.com/eqnedit.php?latex=2^{13}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?2^{13}" title="2^{13}" /></a>

Now - Aristotle can afford to lose 1, 2, 3, 4 or 5 games and still win.

<a href="http://www.codecogs.com/eqnedit.php?latex={13\choose1}&space;+&space;{13\choose2}&space;+&space;{13\choose3}&space;+&space;{13\choose4}+&space;{13\choose5}&space;+&space;1&space;=&space;2380&space;\\&space;\\&space;P(Aristotlte\&space;wins)&space;=&space;\frac{2380}{8192}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?{13\choose1}&space;+&space;{13\choose2}&space;+&space;{13\choose3}&space;+&space;{13\choose4}+&space;{13\choose5}&space;+&space;1&space;=&space;2380&space;\\&space;\\&space;P(Aristotlte\&space;wins)&space;=&space;\frac{2380}{8192}" title="{13\choose1} + {13\choose2} + {13\choose3} + {13\choose4}+ {13\choose5} + 1 = 2380 \\ \\ P(Aristotlte\ wins) = \frac{2380}{8192}" /></a>

= 595/2048

(Almost) exactly what I did.

Now, someone will typically ask why you consider 13 games for each case when 13 games are not played.
Does anyone have that issue?

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

(Almost) exactly what I did.

Now, someone will typically ask why you consider 13 games for each case when 13 games are not played.
Does anyone have that issue?

Yeah and also i tried probability version and it didnt work?

 

[braintic]

Re: 2015 permutation X2 marathon

What are 'counting techniques', because I solved this question through considering cases.

I just meant that the only probability calculation should be the final calculation.

 

[braintic]

Re: 2015 permutation X2 marathon

Yeah and also i tried probability version and it didnt work?

Sorry - my dinner is ready. Perhaps someone else can jump in and answer.

 

[faisalabdul16]

Re: 2015 permutation X2 marathon

(Almost) exactly what I did.

Now, someone will typically ask why you consider 13 games for each case when 13 games are not played.
Does anyone have that issue?

You can explain it - i suck at explaining hahahahaha

 

[RealiseNothing]

Re: 2015 permutation X2 marathon

Doing it my way gives a different answer?

Might be doing something wrong subtly.

 

[RealiseNothing]

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:



This isn't a multiple of 595 though so...

 

[Drsoccerball]

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:



This isn't a multiple of 595 though so...

You're not including 0 L's 1L's etc...

 

[RealiseNothing]

Re: 2015 permutation X2 marathon

Taking the extreme case where Socrates wins 5 consecutive games (putting him in the lead of 9 games to 2), this means that for Aristotle to win, he must win the next 8 games. What does this mean? That Aristotle will win in either 13 or less games.

The total number of outcomes is <a href="http://www.codecogs.com/eqnedit.php?latex=2^{13}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?2^{13}" title="2^{13}" /></a>

Now - Aristotle can afford to lose 1, 2, 3, 4 or 5 games and still win.

<a href="http://www.codecogs.com/eqnedit.php?latex={13\choose1}&space;+&space;{13\choose2}&space;+&space;{13\choose3}&space;+&space;{13\choose4}+&space;{13\choose5}&space;+&space;1&space;=&space;2380&space;\\&space;\\&space;P(Aristotlte\&space;wins)&space;=&space;\frac{2380}{8192}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?{13\choose1}&space;+&space;{13\choose2}&space;+&space;{13\choose3}&space;+&space;{13\choose4}+&space;{13\choose5}&space;+&space;1&space;=&space;2380&space;\\&space;\\&space;P(Aristotlte\&space;wins)&space;=&space;\frac{2380}{8192}" title="{13\choose1} + {13\choose2} + {13\choose3} + {13\choose4}+ {13\choose5} + 1 = 2380 \\ \\ P(Aristotlte\ wins) = \frac{2380}{8192}" /></a>

= 595/2048

Either I'm really derping with this question but surely the total outcomes isn't ?