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. :)
