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

Guest Mar 20, 2015

Best Answer 

 #3
avatar+92191 
+10

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.

Melody  Mar 21, 2015
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4+0 Answers

 #2
avatar+92191 
+5

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)

I ANSWERED THE WRONG QUESTION HERE

THIS IS THE ANSWER TO

6 IDENTICAL B***S AND 3 DIFFERENT BOXES  :  

I HAVE ANSWERED THE ACTUAL QUESTION IN MY NEXT POST

 

0,0,6     the 6 could go in any box so that is  1*3 = 3

0,1,5         3!    permutations here                       =  6

0,2,4         3!    permutations here                       =  6

0,3,3       the 0 could go in any box so that is  1*3 = 3

1,1,4        the 4 could go in any box so that is  1*3 = 3

1,2,3         3!    permutations here                       =  6

2,2,2       there is only one possibility here            =  1

 

 3+6+6+3+3+6+ 1 =  28 ways      

Melody  Mar 21, 2015
 #3
avatar+92191 
+10
Best Answer

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.

Melody  Mar 21, 2015
 #4
avatar+85624 
+5

I might have done it slightly differently, but I believe your answers are correct, Melody....

Here, the boxes don't matter...one is as good as the other...!!!   The only thing  that matters is choosing which b***s to group together in one (or more) boxes. 

Another 3 points from me....!!!

 

  

CPhill  Mar 21, 2015
 #5
avatar+92191 
0

Thanks chris :)

Melody  Mar 21, 2015

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