HSC 2015 MX2 Permutations & Combinations Marathon (archive) (6 Viewers)

porcupinetree · Aug 9, 2015

Re: 2015 permutation X2 marathon

Here's one I just came up with:
A musician is selecting and preparing songs for his upcoming album. He currently has 13 songs: 4 are slow, 6 are moderate, and 3 are fast. How many different (ordered) track listings are possible if:

The album must open and close with a fast song, and
There must be at least 4 moderate and exactly 2 slow songs, and
The album is to contain 9 songs.

Edit: I changed the question to make it a bit easier

Answer: 3810240 (I think)

InteGrand · Aug 9, 2015

Re: 2015 permutation X2 marathon

mreditor16 said:
Your method gets an answer of 960.

OK. Was that the correct answer?

mreditor16 · Aug 9, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
OK. Was that the correct answer?

I actually don't know the correct answer. But I had a go, and got the same. So I think it is

Ekman · Aug 9, 2015

Re: 2015 permutation X2 marathon

porcupinetree said:
Here's one I just came up with:
A musician is selecting and preparing songs for his upcoming album. He currently has 17 songs: 5 are slow, 9 are moderate, and 3 are fast. How many different (ordered) track listings are possible if:

The album must open and close with a fast song, and
There must be at least 3 moderate and 2 slow songs, and
The album is to contain 9 songs.

Edit: I think my answer was wrong, let me think again

I got 152409600

My logic was:
3C1 * 2C1 for the first and last song being a fast one.
7! to arrange the remaining songs in-between the first and last.
Then i let 3 songs be moderate and 2 songs be slow, and considered cases for the remaining two songs. The cases were:
Case 1: The remaining two songs are moderate (Therefore it will be 9C5 * 5C2)
Case 2: The remaining two songs are slow (Therefore it will be 9C3 * 5C4)
Case 3: The remaining two songs are slow and moderate (Therefore it will be 9C4 * 5C3)
Case 4: The remaining two songs are slow and fast (Therefore it will be 9C3 * 5C3)
Case 5: The remaining two songs are moderate and fast. (Therefore it will be 9C4 * 5C2)

Add the cases and multiply by 6 * 7!

porcupinetree · Aug 9, 2015

Re: 2015 permutation X2 marathon

Ekman said:
I got 152409600

My logic was:
3C1 * 2C1 for the first and last song being a fast one.
7! to arrange the remaining songs in-between the first and last.
Then i let 3 songs be moderate and 2 songs be slow, and considered cases for the remaining two songs. The cases were:
Case 1: The remaining two songs are moderate (Therefore it will be 9C5 * 5C2)
Case 2: The remaining two songs are slow (Therefore it will be 9C3 * 5C4)
Case 3: The remaining two songs are slow and moderate (Therefore it will be 9C4 * 5C3)
Case 4: The remaining two songs are slow and fast (Therefore it will be 9C3 * 5C3)
Case 5: The remaining two songs are moderate and fast. (Therefore it will be 9C4 * 5C2)

Add the cases and multiply by 6 * 7!

Looks good to me. I edited/changed the question to make it a bit easier once I realised how many cases there were haha, but your logic seems correct. Here's what I figured for my edited question (in white so that others don't have to see if they don't want):

Case 1 - 4 moderate songs
3C2 . 2! (choose the fast songs to open and close the album)
6C4 (choose the 4 moderate songs)
4C2 (choose the 2 slow songs)
We need to choose 1 more song. We have 1 fast song left, 0 slow songs (exactly 2 slow songs to be in the album) and 0 moderate songs (in this case I chose there to be exactly 4 in the album). Hence 1C1 ways to choose last song.
7! to arrange the middle 7 songs.

Hence total for case 1: 3C2 . 2! . 6C4 . 4C2 . 1C1 . 7! = 2 721 600

Case 2 - 5 moderate songs
3C2 . 2! (same as above)
6C5 (5 mod. songs)
4C2 (2 slow songs)
We already have 9 songs, don't need to pick any more.
7! - arrange

Total for case 2: 1 088 640

Total: 2721600+1088640 = 3810240

kawaiipotato · Aug 9, 2015

Re: 2015 permutation X2 marathon

11 men are waiting at a barber. Three men in particular, A,B,C do not want to sit next to each other. How many ways are possible?

answer: 20321280

Drsoccerball · Aug 9, 2015

Re: 2015 permutation X2 marathon

kawaiipotato said:
11 men are waiting at a barber. Three men in particular, A,B,C do not want to sit next to each other. How many ways are possible?

answer: 20321280

Are you sure the answer is right ?
$11!-9!\cdot 3!$ is my answer

InteGrand · Aug 9, 2015

Re: 2015 permutation X2 marathon

$The answer should be $8!\times {^9\text{P}_3$}, by the insertion method (which is the answer given by kawaiipotato).$

Drsoccerball · Aug 9, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
$The answer should be $8!\times {^9\text{P}_3$}, by the insertion method (which is the answer given by kawaiipotato).$

Amount of combinations unrestricted = 11!
Amount of combinations with A,B,C together = 9!3!
Subtracting the two would give you the amount of times they wont sit next to each other

InteGrand · Aug 9, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
Amount of combinations unrestricted = 11!
Amount of combinations with A,B,C together = 9!3!
Subtracting the two would give you the amount of times they wont sit next to each other

The Q meant that we can't have things like AB together either (with C somewhere else), whereas the method you suggested would fail to exclude these cases.

kawaiipotato · Aug 9, 2015

Re: 2015 permutation X2 marathon

Five women and four men are to be seated at a round table.
(i)
In how many ways may this be done without restrictions?

answer: 40320
(ii)
In how many ways may this be done if no two men are to be seated
together?
answer: 2880
(iii)
If one man and one woman are a married couple, what is the
probability that they are seated together, given
the conditions of part
(ii)?
answer: 2/5

Drsoccerball · Aug 9, 2015

Re: 2015 permutation X2 marathon

kawaiipotato said:
Five women and four men are to be seated at a round table.
(i)
In how many ways may this be done without restrictions?

answer: 40320
(ii)
In how many ways may this be done if no two men are to be seated
together?
answer: 2880
(iii)
If one man and one woman are a married couple, what is the
probability that they are seated together, given
the conditions of part
(ii)?
answer: 2/5

i) 8!
ii)5!4! (arranging them seperately)
iii) $\frac{4!3!(4C1)2!}{2880}$ choosing one of the four couples, fixing them and thus they can switch so 2!, 4!,3! arranging the rest.
PS: Its funny how you had to mention that it was a girl and boy that was married...

braintic · Aug 9, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
PS: Its funny how you had to mention that it was a girl and boy that was married...

Of course, there would be nothing wrong if that were not the case, right?

braintic · Aug 9, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
iii) $\frac{4!3!(4C1)2!}{2880}$ choosing one of the four couples, fixing them and thus they can switch so 2!, 4!,3! arranging the rest.

It is a LOT easier than that.
Each man sits next to 2 of the 5 women. So the probability is 2/5.

kawaiipotato · Aug 11, 2015

Re: 2015 permutation X2 marathon

Find the number of selections of 4 letters from the word APPLES
answer: 11
b) How many 4 letter words can be made from the letters of the word APPLES
answer: 192

rand_althor · Aug 11, 2015

Re: 2015 permutation X2 marathon

kawaiipotato said:
Find the number of selections of 4 letters from the word APPLES
answer: 11
b) How many 4 letter words can be made from the letters of the word APPLES
answer: 192

a) If there are two Ps, we choose 2 letters from the remaining 4, so there are 4C2 selections. If there is one P, we choose 3 letters from the remaining 4, so there are 4C3 selections. 4C2 + 4C3 = 10 selections in total.

b) If two Ps: From a), there are 4C2 selections with two Ps. These can be arranged in (4C2*4!)/2!=72 ways. If one P: From a), there are 4C3 selections with one P. These can be arranged in 4C3*4!=96 ways. In total there are 96+72=168 words.

InteGrand · Aug 11, 2015

Re: 2015 permutation X2 marathon

rand_althor said:
a) If there are two Ps, we choose 2 letters from the remaining 4, so there are 4C2 selections. If there is one P, we choose 3 letters from the remaining 4, so there are 4C3 selections. 4C2 + 4C3 = 10 selections in total.

b) If two Ps: From a), there are 4C2 selections with two Ps. These can be arranged in (4C2*4!)/2!=72 ways. If one P: From a), there are 4C3 selections with one P. These can be arranged in 4C3*4!=96 ways. In total there are 96+72=168 words.

You forgot the case of ''there are 0 P's''.

rand_althor · Aug 11, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
You forgot the case of ''there are 0 P's''.

Ah right thanks, I was wondering where I went wrong.

a) If 0 Ps, then there is only 1 selection with the remaining 4 letters. The total selections is 11.

b) If 0 Ps, then there is 4!=24 ways of arranging the letters. The total amount of words is 192.

Drsoccerball · Aug 12, 2015

Re: 2015 permutation X2 marathon

rand_althor said:
Ah right thanks, I was wondering where I went wrong.

a) If 0 Ps, then there is only 1 selection with the remaining 4 letters. The total selections is 11.

b) If 0 Ps, then there is 4!=24 ways of arranging the letters. The total amount of words is 192.

$5P4 + \frac{(4C2)4!}{2!}$

braintic · Aug 12, 2015

Re: 2015 permutation X2 marathon

In how many ways can you climb 12 stairs by stepping 1 or 2 steps at a time?
(Not necessarily all 1's or all 2's)
Answer: 233

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