PLZ Help ME!!!FAST

Hi Bobo2k16, it is very nice to meet you :)

1. How ways are there to put 6 b***s in 3 boxes if the b***s are not distinguishable but the boxes are?

ok boxes are different but ball are the same.  Mmm do I have to put at least one ball in every box.  I guess some boxes can be empty...

So the only relevent thing is how many b***s are in each individual box.

I think I wil use the stars and bars method.

I'l put the boxes in order   A,B,C

now I have the b***s

1,2,3,4,5,6    they are the stars, now I wil use 2 bars to divide those b***s into three groups.

So now there are 6 stars and 2 bars and I want to know how many places in the line I can put the 2 bars.

That will be 8C2=28

So that is 28 ways :)

Here is a clip to explain the stars and bars method to you

https://www.youtube.com/watch?v=5hF59P8tmQc

2.How many ways are there to put 6 b***s in 3 boxes if the b***s are distinguishable but the boxes are not?

THX!!!

Mmm that is harder...

the 3 boxes are all the same but the 6 b***s are distinguishable...  

Boxes A,B,C but they are the same

6,0,0

5,1,0

4,2,0

4,1,1

3,3,0

3,2,1

ok there are 6 ways the b***s can be put into the boxes by number but that doesn't take into account that the b***s are different from each other....

6,0,0      all in the same box so 1 way      

5,1,0     there are 6 to chose from to go into the second box so that is 6 ways   6C1

4,2,0     6C2 ways = 15

4,1,1

3,3,0     6C3 ways = 20

3,2,1

Now the last 2 that I haven't done yet are tricky.  I severely risk multiple counting here.  Mmm

4,1,1   There are 6C4 ways to choose the first one and then it doesn't matter which box the others go into since the boxes are the same so that is 6C4*1 = 6C4=15

3,2,1  the trickiest one...      that would be  6C3*3C1=6C3*3= 20*3=60     I think but I am not totally confident.

So I have

1+6+15+20+15+60 = 117 ways

Bobo2k16,

It would be nice if you commented on the answer.

Answerers like to know that the asker actually reads and at least tries to comprehend their answers  !!

1+6+15+20+15+60 = 117 ways

It has been bought to my attention that I forgot the combination 2,2,2

There are 15 combinations here so that makes   117+15 = 132 ways

Hi all,

I thought I would give myself an alter ego.  Only becasue I want people to know that Bertie still has some input sometimes and I am really pleased about that.  People who have been here a long time will remember Bertie.  He is the one, on this forum, who is best at these types of questions.  For some reason he can't log in anymore and has consequently stopped coming here :(

Anyway....

I asked Bertie to check this question for me.  He says it is correct, just that I forgot the 2-2-2 combination.

I already fixed that in answer #2.

I also asked him to discuss how you can derive this answer for the 2-2-2 scenario.  I could muffle through it and get the correct answer but I wanted Bertie to discuss it with me so I could do it more confidently next time.

Here is his response :)

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Hi Melody,
I didn't use a formula for the 2-2-2 case, I sort of muddled my way through it, (my usual method).
I thought in terms of six objects, call them A B C D E F and formed them into three pairs.
A has to paired with any one of the other five and that will naturally give rise to five groups, headed by AB, AC, AD, AE and AF
Concentrating on the group headed by AB, the C will be paired with any one of the other three, so we have CD, CE and CF
The final pair takes care of itself.
That means that the group headed AB, will contain three entries, 
AB CD EF, AB CE DF, AB CF DE.
There are five groups in all, so 5*3*1 = 15 possibilities.

I asked Bertie if he could extend the logic to similar questions with more b***s:

Follow that logic for the eight objects and four boxes and you arrive at the result 7*5*3*1 = 105.

From that, it's possible to deduce a formula. (I seem to remember doing something like this awhile ago ?)
7*5*3*1 = 8*7*6*5*4*3*2*1/(8*6*4*2) = 8!/(2^4*4!),
and in general,
(2n-1)*(2n-3)*(2n-5)*...*5*3*1 = (2n)!/(2^n*n!).

For the six object case that we started with, we would have, with n = 3,
6!/(2^3*3!) = 720/(8*6) = 15.

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I really like Bertie's alternate solution that follows!   It makes perfect sense and I found it easy to comprehend.  

That also suggests an alternative way of looking at the solution.

Consider the six object case.

The six objects can be arranged into 6! = 720 different orderings.
Take the first one, ABCDEF say, and split it, as it is, into three pairs AB CD EF.
So far as the problem is concerned, there will be many repetitions of this, BA CD EF,
DC AB EF, FE BA CD for example etc.
The order of the objects in each pairing could be reversed and/or the order of the pairs themselves could be different. The repetitions would be removed by dividing by 2^3 to remove the ordering of the objects in each pair, and by 3! to remove the ordering of the pairs. That gets us to the earlier result 6!/(2^3*3!).

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I know that Bobo (the question asker)  has said part B is incorrect and maybe he is right but it may pay to remember that Bobo is a clown. (sorry I could not resist)    Bobo has not indicated why he thinks the answer is wrong.  Maybe the answer in his book is incorrect.

I will put this question in with our resource proability questions which you can find via the sticky topics.   

If you see this post Bertie, I am very grateful for your help.  :)