Number of Arrangements?

[asa.hoshi]

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 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?

[kamil9876]

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 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 ways of doing this.

So in total there are ways.

[asa.hoshi]

Thanks. but am i see something, but does this way allows GBGBGBB when you shouldn't (in the 1st section)?

[kamil9876]

actually yeah fail.

[asa.hoshi]

actually yeah fail.

u gave a better attempt than i did! HAHA.

[kamil9876]

I have a solution but it involves some casework, so i will wait and see if there is a better one.

[asa.hoshi]

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.

[kamil9876]

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.

[asa.hoshi]

i think your way is more efficient.

hey at least we solved the problem 

[kamil9876]

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 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

[dcc]

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.

[kamil9876]

1260-79=1181

Who are "they" btw? I'm curious

[asa.hoshi]

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?

[kamil9876]

guess so, would like to see a better method if possible.

What subject is this though?

[asa.hoshi]

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.