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

#2**+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

#1**+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

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**+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