HSC 2015 MX2 Permutations & Combinations Marathon (archive) (5 Viewers)

[Drsoccerball]

Since there's an individual thread for integration there should be one for something far more challenging.
Rules:
1) Answer questions before adding more
2) Questions must have the answer(Since permutations are trivial) but write them in white as a way of checking!!
3) Non-2015ers must wait and not answer a question without giving the 2015'ers the chance...
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A zoo keeper wants to place 6 tigers in 4 cages, each can fit a maximum of two tigers.
In how many ways can he place the tigers if no cage to be left empty?
Answer: 1080

 

[InteGrand]

Re: 2015 permutation X2 marathon

Since there's an individual thread for integration there should be one for something far more challenging.
Rules:
1) Answer questions before adding more
2) Questions must have the answer(Since permutations are trivial)
3) Non-2015ers must wait and not answer a question without giving the 2015'ers the chance...
------------------------------------------------------------------------------------------------------------------------------
A zoo keeper wants to place 6 tigers in 4 cages, each can fit a maximum of two tigers.
In how many ways can he place the tigers if no cage to be left empty?
Answer:1080

How are permutations and combinations Q's trivial in general? Also, we should probably write the answer in white font or something, so we don't see the answer straightaway. Seeing the answer makes the question much much easier.

 

[braintic]

Re: 2015 permutation X2 marathon

I wouldn't have a clue how to write the answers in white.

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

I wouldn't have a clue how to write the answers in white.

When the quick reply box comes up theres an "A" click on it and pick the white font

 

[braintic]

Re: 2015 permutation X2 marathon

When the quick reply box comes up theres an "A" click on it and pick the white font

How do you get that font that shows up when you roll your mouse over it? This is what I thought you meant.

 

[rand_althor]

Re: 2015 permutation X2 marathon

How do you get that font that shows up when you roll your mouse over it? This is what I thought you meant.

Do you mean spoilers? e.g

Spoiler

abc

For that the text you want to be hidden is enclosed in SPOILER tags so [ SPOILER]text here[/ SPOILER] (without the spaces)

 

[braintic]

Re: 2015 permutation X2 marathon

Do you mean spoilers? e.g

Spoiler

abc

For that the text you want to be hidden is enclosed in SPOILER tags so [ SPOILER]text here[/ SPOILER] (without the spaces)

Thanks - I'll try to remember that.

 

[InteGrand]

Re: 2015 permutation X2 marathon

The problem is, the spoilers don't work well here, they don't block the text UNTIL you scroll your mouse over it. The white font works better.

 

[porcupinetree]

Re: 2015 permutation X2 marathon

A zoo keeper wants to place 6 tigers in 4 cages, each can fit a maximum of two tigers.
In how many ways can he place the tigers if no cage to be left empty?
Answer: 1080

Here's my attempt, it's double your answer but I can't figure out why

 

[braintic]

Re: 2015 permutation X2 marathon

The white font works better.

IF you can locate it. Perhaps the word "Answer" should not be in white font - just the answer itself.

 

[InteGrand]

Re: 2015 permutation X2 marathon

IF you can locate it. Perhaps the word "Answer" should not be in white font - just the answer itself.

Yeah that's what I was intending.

 

[Soulful]

Re: 2015 permutation X2 marathon

Since there's an individual thread for integration there should be one for something far more challenging.
Rules:
1) Answer questions before adding more
2) Questions must have the answer(Since permutations are trivial) but write them in white as a way of checking!!
3) Non-2015ers must wait and not answer a question without giving the 2015'ers the chance...
------------------------------------------------------------------------------------------------------------------------------
A zoo keeper wants to place 6 tigers in 4 cages, each can fit a maximum of two tigers.
In how many ways can he place the tigers if no cage to be left empty?
Answer: 1080

The only "formation" that fulfills the restrictions, namely, two cages with two tigers, and two cages with 1

Thus, we first choose 2 cages - 4C2
For the 1st cage, we choose 2 tigers - 6C2
For the 2nd cage, we choose another 2 tigers - 4C2
There are only two ways to slot the remaining 2 tigers

Total = 4C2 x 6C2 x 4C2 x 2 = 1080

next question

how many ways can you arrange COMBINATION so that the vowels are in alphabetical order and separated from each other by at least 1 consonant

Answer: 7560

 

[Soulful]

Re: 2015 permutation X2 marathon

Here's my attempt, it's double your answer but I can't figure out why

When you choose two cages for the other two tigers, you count cases where the two tigers end up in the same cage, meaning there will be a cage with 3 tigers (which goes against the restrictions)

 

[kawaiipotato]

Re: 2015 permutation X2 marathon

The only "formation" that fulfills the restrictions, namely, two cages with two tigers, and two cages with 1

Thus, we first choose 2 cages - 4C2
For the 1st cage, we choose 2 tigers - 6C2
For the 2nd cage, we choose another 2 tigers - 4C2
There are only two ways to slot the remaining 2 tigers

Total = 4C2 x 6C2 x 4C2 x 2 = 1080

next question

how many ways can you arrange COMBINATION so that the vowels are in alphabetical order and separated from each other by at least 1 consonant

Answer: 7560

is this valid? if not, why?

A ** I ** I ** O ** O **
where the stars are the possible ways letters can go into (C,M,B,N,N,T)

Since there's 10 spots for letters to go into: 10C6
And since you can shift the vowels as a group until the last O touches the last star = 3
So total = 10C6 x 3 = 630

 

[porcupinetree]

Re: 2015 permutation X2 marathon

When you choose two cages for the other two tigers, you count cases where the two tigers end up in the same cage, meaning there will be a cage with 3 tigers (which goes against the restrictions)

I think I figured out another way to fix mine:

6C4 (choose 4 tigers and put in cages)
4! (arrange these tigers)
4C2 (choose 2 cages for other two tigers)
2! (ways to arrange these two tigers)
divided by 2!2!, because in each of the two 2-tiger cages, there are 2! ways to arrange the tigers, but these arrangements are effectively identical.
=1080

 

[rand_althor]

Re: 2015 permutation X2 marathon

is this valid? if not, why?

A ** I ** I ** O ** O **
where the stars are the possible ways letters can go into (C,M,B,N,N,T)

Since there's 10 spots for letters to go into: 10C6
And since you can shift the vowels as a group until the last O touches the last star = 3
So total = 10C6 x 3 = 630

It is not correct. The correct answer is in white under the question. For your calculation, what if the 6 consonants become the first 6 stars? Then we could have A C M I B N I N T O O. For this case, the last two vowels are not separated by a consonant.

 

[sida1049]

Re: 2015 permutation X2 marathon

The only "formation" that fulfills the restrictions, namely, two cages with two tigers, and two cages with 1

Thus, we first choose 2 cages - 4C2
For the 1st cage, we choose 2 tigers - 6C2
For the 2nd cage, we choose another 2 tigers - 4C2
There are only two ways to slot the remaining 2 tigers

Total = 4C2 x 6C2 x 4C2 x 2 = 1080

next question

how many ways can you arrange COMBINATION so that the vowels are in alphabetical order and separated from each other by at least 1 consonant

Answer: 7560

Consider the consonants as "fixed" positions:
_C_C_C_C_C_C_
Where the underscores highlight the positions the vowels can situate.

Hence the number of possible positions of vowels follows: 7C5

Vowels can only be in one arrangement.

There are 6 consonants but 2 are of the same, hence number of arrangement of vowels = 6!/2

Answer: 7C5.6!/2 = 7560.

Next question:

There are 8 gates which 10 cars can pass through. How many ways can the cars pass through if every gate is used?(Order of cars passing through a particular gate does not matter, but order of cars passing through different gates, e.g. car A through gate 1 or gate 2, matters.)

(Sorry my initial answer was incorrect. Working out another one.)

Oh my god I've just realised the difficulty of this question. Still working on it. Apologies for the edits.

 

[kawaiipotato]

Re: 2015 permutation X2 marathon

Consider the consonants as "fixed" positions:
_C_C_C_C_C_C_
Where the underscores highlight the positions the vowels can situate.

Hence the number of possible positions of vowels follows: 7C5

Vowels can only be in one arrangement.

There are 6 consonants but 2 are of the same, hence number of arrangement of vowels = 6!/2

Answer: 7C5.6!/2 = 7560.

Next question:

There are 8 gates which 10 cars can pass through. How many ways can the cars pass through if every gate is used?

(Sorry my initial answer was incorrect. Working out another one.)

how come we don['t account for when it's like _C_C_C_C_C_CC_ ?

 

[Ekman]

Re: 2015 permutation X2 marathon

Consider the consonants as "fixed" positions:
_C_C_C_C_C_C_
Where the underscores highlight the positions the vowels can situate.

Hence the number of possible positions of vowels follows: 7C5

Vowels can only be in one arrangement.

There are 6 consonants but 2 are of the same, hence number of arrangement of vowels = 6!/2

Answer: 7C5.6!/2 = 7560.

Next question:

There are 8 gates which 10 cars can pass through. How many ways can the cars pass through if every gate is used?

(Sorry my initial answer was incorrect. Working out another one.)

For the gates question, my plan of attack was:

10C8 different cars to go through every single gate
8! for each gate to be accessed
8^2 for the remaining two cars, since we don't care which gate they access

10C8 * 8! * 8^2 = 116121600

 

[sida1049]

Re: 2015 permutation X2 marathon

For the gates question, my plan of attack was:

10C8 different cars to go through every single gate
8! for each gate to be accessed
8^2 for the remaining two cars, since we don't care which gate they access

10C8 * 8! * 8^2 = 116121600

There is another case where 3 cars go through a single gate and 7 gates with a single car.