ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: asa.hoshi on October 03, 2010, 04:54:51 pm

Title: Number of Arrangements?
Post by: asa.hoshi on October 03, 2010, 04:54:51 pm: Calculate the number of arrangements of 2 red, 3 green and 4 blue bottles in a line, given that at least 2 bottles of the same colour are always to be in succession.

I know that...
#arrangements w/o restrictions is $\frac{9!}{2!3!4!}=1260$ ways

need to find # arrangements so at least 2 bottles of the same colour are in succession.
So,

if i take #arrangements w/o restrictions - #arrangements with no colour in succession,

would that work out? but im having heaps of toruble to count #arrangements with no colour in succession... hrm, is there another way to approach this question?
Title: Re: Number of Arrangements?
Post by: kamil9876 on October 03, 2010, 08:45:57 pm: Yes, that's a good start.

To solve our next little sub-problem, I'll show you that it is easier if we assume there are only two kinds of bottles, ie let us forget about the red.

So we have:

_ B _ B _ B _ B _

and we must choose exactly 3 of the _ to place our G.

So there are $\binom{5}{3}$ different ways of doing this.

Now how do I also include the fact that there are 2 red bottles?

Well for any arrangement of the 7 non-red bottles looks like this:

_ X _ X _ X _ X _ X _ X _ X _ X _

Where X denotes any of the non-red bottles. We know how many ways there are of arranging the X's, and now we know that for each arrangement we must choose exactly 2 of the _ to place the red bottles. There are $\binom{8}{2}$ ways of doing this.

So in total there are $\binom{5}{3} \binom{8}{2}$ ways.
Title: Re: Number of Arrangements?
Post by: asa.hoshi on October 03, 2010, 11:41:39 pm: Thanks. but am i see something, but does this way allows GBGBGBB when you shouldn't (in the 1st section)?
Title: Re: Number of Arrangements?
Post by: kamil9876 on October 04, 2010, 12:37:53 am: actually yeah fail.
Title: Re: Number of Arrangements?
Post by: asa.hoshi on October 04, 2010, 12:46:41 am: Quote from: kamil9876 on October 04, 2010, 12:37:53 am
actually yeah fail.
u gave a better attempt than i did! HAHA.
Title: Re: Number of Arrangements?
Post by: kamil9876 on October 04, 2010, 12:00:28 pm: I have a solution but it involves some casework, so i will wait and see if there is a better one.
Title: Re: Number of Arrangements?
Post by: asa.hoshi on October 04, 2010, 05:39:31 pm: haha.thanks for your help. but i kinda solved it. i used ur idea, _B_B_B_B_
and then placed the Rs and counted the possible ways to place the Gs w/o colour succession.
i.e. RB_B_B_BR, G must go where the _ and there is only 1 way to do it...
then i went on RBRB_B_B, x2 G must go where the _ are, and the remaining G can go in 5 different spots within the arrangement ect...
I came up with 79. I think its right.
Title: Re: Number of Arrangements?
Post by: kamil9876 on October 04, 2010, 07:07:51 pm: yeah that's the idea I used, I split it into four essentially different cases and the sum was 10 + 3*2*8 + 2*2*3 + 3*3=79.

It wouldn't be so nice for arbitrary number of bottles though, though maybe the problem is too complex for that.
Title: Re: Number of Arrangements?
Post by: asa.hoshi on October 04, 2010, 07:48:18 pm: i think your way is more efficient.

hey at least we solved the problem ;)
Title: Re: Number of Arrangements?
Post by: kamil9876 on October 04, 2010, 08:52:00 pm: if you wanna know the cases were:

X denotes where to place the remaining 5 bottles.

1) X B X B X B X B X

There are $\frac{5!}{3!2!}=10$ ways since u can just ignore the Bs to count.

2) B X X B X B X B X

If X X is G R then there are 3 ways (since the R can go in any remaining X). or if X X is R G then same story, so 2*3=6 ways. However this can be arranged in 8 differents ways like X X B X B X B X B etc. so 2*3*8 altogether.

3) B XX B XX B X B

4) B XXX B X B X B
Title: Re: Number of Arrangements?
Post by: dcc on October 15, 2010, 09:58:29 pm: turns out the real answer was 1181 - and they didn't accept other (in my opinion, reasonable) interpretations of the question. which is a shame.
Title: Re: Number of Arrangements?
Post by: kamil9876 on October 15, 2010, 10:45:12 pm: 1260-79=1181

Who are "they" btw? I'm curious
Title: Re: Number of Arrangements?
Post by: asa.hoshi on October 16, 2010, 10:37:43 pm: i think he was refering to the subject lecturer and tutor. that was an assignment problem. and the answer was indeed 1181. lol. guess we got it right yeh kamil9876?
Title: Re: Number of Arrangements?
Post by: kamil9876 on October 17, 2010, 12:39:26 am: guess so, would like to see a better method if possible.

What subject is this though?
Title: Re: Number of Arrangements?
Post by: asa.hoshi on October 17, 2010, 05:15:38 pm: Discrete Mathematics is the subject. I think how the tutor did it was similar to how u did it...lol. the lecturer is yet to post the solutions up for that assignment. so i'll let you know how the lecturer does it after he posts up the solutions.
Title: Re: Number of Arrangements?
Post by: kamil9876 on October 17, 2010, 05:19:31 pm: o ok with Richard Brak lol?

I will most likely do this next year, pity I didn't realise that 2nd year discrete math and OP is not neccesary for 3rd year discrete math. Then I wouldn't have had to do that crap and could have done this instead.