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

tertre Sep 17, 2016

#4**+15 **

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

CPhill Sep 17, 2016

#1**+5 **

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

Melody Sep 17, 2016

#4**+15 **

Best Answer

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

CPhill Sep 17, 2016