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

#3**+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

#2**+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**+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**+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