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# Distinguishability

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

I could use some help!

Aug 25, 2018
edited by Guest  Aug 25, 2018

#9
+100529
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Here's my attempt at this....

Since the boxes are indistinguishable, it really just boils down to the selection of balls to put in the first box...also...I'm assuming that we can have empty boxes

[ I think Melody's assumption is slightly incorrect  ...the boxes are the same but the balls are different ]

4 balls in the first box  =  1 way

3 balls in the first box....the other ball in either two of the remaining boxes  = C(4,3)  =  4 ways

2 balls in the first box and 2 others in any remaining box  = C(4,2)  / 2  =

6 /2  = 3 ways

{  The reason for dividing by 2 is that  if we selected balls 1 and 3  for the first box, balls 2 and 4 would occupy either of the other two boxes....but...this is the same as  selecting balls 2 and 4 for the first box and balls 1 and 3 to occupy either of the other two boxes )

2 balls in the first box and  one in each of the other two  = C(4,2)  = 6 ways

So...by my count....the total ways  =  1 + 4 + 3 + 6  =  14 ways

Aug 26, 2018
edited by CPhill  Aug 26, 2018

#1
+1

Go online to this site and use the formula in the table at the bottom under "Distinct,  identical"

Permutations and Combinations > Distributing Balls into Boxes

Aug 25, 2018
#2
+1

I did that, but what does the "S(k,i) mean?

Aug 25, 2018
#4
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Understanding Sterling numbers isn't too difficult. But going to the site is very difficult, and it's impressive you went there without a link.

Guest Aug 25, 2018
#3
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S(k, i)=Stirling Numbers of the Second Kind. They are computed using "Summation Formula" as explained at the bottom of their table. Here is the Summation formula:

1/3!*∑[(-1)^i * 3 nCr i * (3 - i)^4, i, 0, 3] =6, where 3= Number of boxes, n=Number of balls.

Note: If you don't understand it, I sympathize with you. Your teacher should explain it to you in detail.

Aug 25, 2018
#5
+100813
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How many ways are there to put 4 balls in 3 boxes if the balls are distinguishable but the boxes are not?

The  3 boxes are all different and the 4 balls are identical

All 4 be in any box   That is 3 ways

3 in one box and 1 in another that would be   3*2 = 6 ways

2 in one box and 2 in another that would be   3*2 = 6 ways

2 in one box and 1 in each of the others   3 ways

That is a total of 18 ways I think.

Aug 26, 2018
#6
+100813
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No one else actually gave an answer, but if someone wants to compare my answer to the formula answer that would be interesting.

There is always plenty of room to go wrong with probablility but then I don't trust formulas if I do not understand how they are derived. So I would not automatically accept a formula answer or my own answer.  Either or both could be wrong.

Maybe they are the same and they are both right. That would almost be a first!    LOL

Melody  Aug 26, 2018
#7
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Probability is a "curse" upon the world of mathematics!! Here is EXACTLY the same question and the 6 answers given, including one by a Professor of Mathematics, almost all of which differ from each other in their final answer!

https://www.quora.com/In-how-many-ways-can-4-distinct-balls-be-distributed-into-3-identical-boxes

Aug 26, 2018
#8
+100813
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Original post deleted. I would have left it here but it just caused confusion.

Sorry guest.

Oh I will explain that I misread the question so everyone can ignore my answer.

Melody  Aug 26, 2018
edited by Melody  Aug 26, 2018
edited by Melody  Aug 26, 2018
#9
+100529
+2

Here's my attempt at this....

Since the boxes are indistinguishable, it really just boils down to the selection of balls to put in the first box...also...I'm assuming that we can have empty boxes

[ I think Melody's assumption is slightly incorrect  ...the boxes are the same but the balls are different ]

4 balls in the first box  =  1 way

3 balls in the first box....the other ball in either two of the remaining boxes  = C(4,3)  =  4 ways

2 balls in the first box and 2 others in any remaining box  = C(4,2)  / 2  =

6 /2  = 3 ways

{  The reason for dividing by 2 is that  if we selected balls 1 and 3  for the first box, balls 2 and 4 would occupy either of the other two boxes....but...this is the same as  selecting balls 2 and 4 for the first box and balls 1 and 3 to occupy either of the other two boxes )

2 balls in the first box and  one in each of the other two  = C(4,2)  = 6 ways

So...by my count....the total ways  =  1 + 4 + 3 + 6  =  14 ways

CPhill Aug 26, 2018
edited by CPhill  Aug 26, 2018
#10
+100813
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LOL

Chris

"if the balls are distinguishable"     It is the balls that are the same.

edit:

Woops you are right - I kept seeing indistinguishable.

I hate that word, those words, distinguishable and indistinguishable The are totally confusing.

Why can't the question use the words different and identical  !!

I goofed up    Sorry.

Melody  Aug 26, 2018
edited by Melody  Aug 26, 2018
#12
+100529
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Sorry, Melody...but if the balls are distinguishable, we  can tell them apart....so Ball1  is different from Ball 2 ....and those are different from Ball 3...etc.

The boxes are "indistinguishable"....we cannot tell them apart....Box 1  is the same as Box 2....etc...

CPhill  Aug 26, 2018
#14
+100529
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HAHAHA!!!.....that's like the words  "inflammable"  and "flammable"....they both mean the same thing......!!!

The English language is strange, isn't  it  ???

CPhill  Aug 26, 2018
#16
+100813
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Yes, sometimes it is just plain stupid.

inflammable and flamable is one  off the absolute stupidest examples of our language.

We both agree on that!

Another stupid word is 'pitted'.

If a date is pitted, does that mean its pit is taken out or does it mean that the pit is still in it.

OR

Can it mean either one?   Beats me!

Melody  Aug 26, 2018
#17
+100529
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LOL!!!!

CPhill  Aug 26, 2018
#11
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Here is the question on the Internet:

"In how many ways can 4 distinct balls be distributed into 3 identical boxes?"

Isn't the word "distinct" the same as "distinguishable", and the word "identical" the same as "indistinguishable"? If so, then it is exactly the same question.

Aug 26, 2018
#13
+100813
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Yea sorry guest, I goofed up

Melody  Aug 26, 2018
#15
+100813
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ok I will give it another shot

3 identical boxes and 4 different balls. Have I got it right this time?

possible numbers of balls in each of the identical boxes

4,0,0

3,1,0

2,2,0

2,1,1

All four in a box,    1 way

how many ways can i choose 3 balls? well one must be left out so that is 4 ways

So there are 4 ways that 3 balls go into one box and 1 in another.

how many ways can I chose 2 balls.   mmm  4C2= 6 ways

so there are 6 ways that 2 can go in one box and 2 in another.

BUT hang on, ive double counted here, there is really only 3 ways

There are also 6 ways that 2 can go in one box and one in each of the other boxes  6 ways

I get 14 ways.

That is what CPhill got too :)

I think in this instant that Quora answerer is incorrect. They have double counted on the 2-2-0 posibility.

Aug 26, 2018