How many ways are there to put 4 b***s in 3 boxes if the b***s are distinguishable but the boxes are not?

I believe that the answer is 14

We have

4 0 0    =  1 way

3 1 0    =   4 ways  to choose the ball that will occupy one of the boxes

2 2 0    =  C(4,2) / 2 =  3 ways.......to see this......note that we could choose to put b***s 1 and 2 into the frist box and 3 and 4 into the second....but.......we could also choose to put b***s 3 and 4 into the first box and 1 and 2 into the second.......but, since the boxes are indistinguishable, this is actually the same arrangement....so  we must divide C(4,2) by 2 to avoid "double-counting"

2 1 1  =  C(4,2)  = 6 ways........we can choose any 2 of the 4 b***s....the order of placing the remaining two b***s into the other two boxes is irrelevant since the boxes are indistinguishable

So......1 + 4 + 3 + 6 =  14 ways

{Note.... I don't understand this very well, but apparently we can also calculate this by something known as "Sterling Numbers of the Second Kind" }......this  =  S2(4,1) + S2(4,2) + S2(4,3)   =  14

How many ways are there to put 4 b***s in 3 boxes if the b***s are distinguishable but the boxes are not?

You can have

4   0    0        one way

3   1    0        4 ways because any of the 4 b***s can be by itself

2   2    0         4C2 = 6 ways

2   1    1         4C2 = 6 ways

Total 1+4+6+6 = 17 ways

it says its wrong?

melody plz help 

Sorry.....it should be "Stirling"  instead of "Sterling"

thank u so much.................

Thanks Chris,

See my answer #6       

Hey.......where's my "stars???".......LOL!!!!!!

What are you on about Chris,

I gave you 5 stars!!    I put them on the boxes along with the b***s, didn't you see them.    LOL

What is the point in wearing glasses if you can see anyway!!

Here are some new specs for you.     

OK.....since taking off my "sunnies" and putting on the specs you have provided.....I see them clearly, now.....

HAHAHAHAHA!!!!!!!!!!