Melody

Thanks chris :)

6 DIFFERENT B***S AND 3 DIFFERENT BOXES

THIS HAS BEEN EDITED

I BELIEVE CPHILL AND I ARE FINALLY IN TOTAL AGREEMENT

DIFFERENT BOXES - 3 WHITE AND 3 BLACK B***S.

0,0,6             3*1=3  choice

0,1,5              3!* the 1 could but black or white   6*2=12 choices

0,2,4              3!*the 2 could be BB, WW or BW     6*3 = 18 choices

0,3,3              3* the 3 could be  BBB, BBW, BWW or WWW but 2 of these are the same so 2 choices

                      (BBB,WWW)  or   (BBW)(WWB)     3*2= 6 choices

1,1,4              3* the ones can be   BB, WW or BW       3*3 = 9 choices

1,2,3              3!*    

                      If the first box has a B then the second can have BB, BW or WW

                      If the first box has a W then the second can have BB, BW or WW     6*6= 18 choices

2,2,2              1* If one box has WW then the other boxes must be BB and BW.

                     or    BW,BW, BW                                                                   1*2=   2 choices

$${\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{68}}$$

There is no guarantees that this is correct.

Well, I got 7 ways if the boxes are all  the same so I am going to start with those solutions. (I hope I got them all. I think i did)

6 DIFFERENT B***S - 3 IDENTICAL BOXES

0,0,6       1 way

0,1,5        6C1 = 6 ways

0,2,4        6C2 = 15 ways

0,3,3       6C3 = 20 ways

1,1,4        6*5=30 ways

1,2,3        6*5C2 = 60 ways

2,2,2       6C2*4C2 = 15*6=90 ways

$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}}{\mathtt{\,\small\textbf+\,}}{\mathtt{30}}{\mathtt{\,\small\textbf+\,}}{\mathtt{60}}{\mathtt{\,\small\textbf+\,}}{\mathtt{90}} = {\mathtt{222}}$$      

That is what I think but it may not be correct.

Thanks Chris

$$\frac{d}{dx}x^{4/3}=\frac{4}{3}\times x^{1/3}=\frac{4\sqrt[3]{x}}{3}$$

If the b***s are all the same and the boxes are all the same then the only difference is how many b***s are in the boxes.

I counted them and I get 7 ways

0,0,6

0,1,5

0,2,4

0,3,3

1,1,4

1,2,3

2,2,2

Since all the b***s and all the boxes are the same, the only difference can be is how many b***s  are in the boxes.

YES the question does clearly state that - next time I really must wear my glasses when i am reading questions.

I probably misread this same word in my other answers as well which would make them all wrong.  

The joke is on me     ROF LOL

Thanks Chris. 

CPhill's second answer may well be correct - that is certainly how I should have started it but I have not worked it out all through to the end. :)

Well now, that is interesting.     LOL

I am not laughing at Chris, my answer could be wrong.  The 2 extreme answers are funny though.  :)

I got the same.  Miracles to happen sometimes.  LOL