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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!

Guest Aug 25, 2018

edited by
Guest
Aug 25, 2018

#9**+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

#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**

Guest Aug 25, 2018

#3**+1 **

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.

Guest Aug 25, 2018

#5**+2 **

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.

Melody Aug 26, 2018

#6**+2 **

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**+1 **

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**

Guest Aug 26, 2018

#9**+2 **

Best Answer

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

#10**+2 **

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

#12**+1 **

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**+1 **

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**+2 **

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

#11**+1 **

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.

Guest Aug 26, 2018

#15**+3 **

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.

Melody Aug 26, 2018