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

+5
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11
+4330

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

Sep 17, 2016

#4
+106532
+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

Sep 17, 2016

#1
+106943
+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

Sep 17, 2016
#6
+106943
+5

Thanks Chris,

Yes what Chris has said does sound right.

2,2,0   should be   4C2/2 = 3

So I agree with Chris,  I think that there are 14 ways too.

Melody  Sep 17, 2016
#2
+4330
+5

it says its wrong?

Sep 17, 2016
#3
+4330
+5

melody plz help

Sep 17, 2016
#4
+106532
+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
#5
+106532
+5

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

Sep 17, 2016
#7
+4330
+10

thank u so much.................

Sep 17, 2016
#8
+106943
+5

Thanks Chris,

Sep 17, 2016
#9
+106532
+5

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

Sep 17, 2016
#10
+106943
+15

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.

Sep 17, 2016
#11
+106532
+10

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

HAHAHAHAHA!!!!!!!!!!

Sep 17, 2016