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How many ways are there to put 6 b***s in 3 boxes if three b***s are indistinguishably white, three are indistinguishably black, and the boxes are distinguishable?

Guest Mar 20, 2015

Best Answer 

 #2
avatar+91510 
+5

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.

Melody  Mar 21, 2015
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2+0 Answers

 #1
avatar+91510 
+5

How many ways are there to put 6 b***s in 3 boxes if three b***s are indistinguishably white, three are indistinguishably black, and the boxes are distinguishable?

AGAIN i HAVE ANSWERED A DIFFERENT QUESTION HERE.

THIS IS   BOXES ALL THE SAME,   3 BLACK AND 3 WHITE B***S

Again I am basing this one my original answer - if it its wrong then they are all wrong

original answer

 

0,0,6             1 choice

0,1,5              the 1 could but black or white   2 choices

0,2,4              the 2 could be BB, WW or BW      3 choices

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

                      (BBB,WWW)  or   (BBW)(WWB)

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

1,2,3              Now it is getting harder

                      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 choices

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

                     or    BW,BW, BW                                                                             2 choices

 

I hope I have got them all   :/

1+2+3+2+3+6+2 = 19 possibilities

Melody  Mar 21, 2015
 #2
avatar+91510 
+5
Best Answer

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.

Melody  Mar 21, 2015

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