How many ways are there to put 6 b***s in 3 boxes if the b***s are distinguishable and the boxes are distinguishable?

My first answer was c**p....I'll try again

If the b***s and boxes are distinguishable

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We can put all 6 b***s into any box.....there are 3 ways to do this....put them into the first box, the second box or the third box

So 3 ways in this arrangement

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We can put 5 b***s into one box, 1 in another and 0 into the last

 But since the b***s are distinguishable, we can pick any 5 of the 6 to put into a box and we have 3 ways to choose that box. =  3C(6,5) = 18 ways

And we have 1 way to choose the next ball and 2 ways to select the next box = 2 ways

The last box is empty and determined by default

So 18 x 2 = 36 ways in this arrangement

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And we can put 4 b***s into one box and 1 in each of the remaining two.

So we have 3 ways to choose the first box and we want to select any 4 of the 6 b***s to put into that box. So 3C(6,4) = 45 ways

And we have 2 ways to choose the next box and 2 ways to choose the next ball = 4 ways

And the final box and ball are determined by default

So 45 x 2 = 180 ways in this arrangement

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And we can select 3 b***s in one box, 2 into another and 1 in the last

We have 3 ways to select the first box and we want to choose 3 of the 6 b***s to put into that box......so we have 3C(6,3)  = 60 ways to do that

Then, we have 2 ways to choose the next box to put 2 of the remaining 3 b***s into....so this is  2C(3,2)  = 6 ways

Again the last ball and and box are determined by default

So the total ways here are  60 x 6 = 360 ways in this arrangement

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And we can put 4 b***s into one box and two into another

We have 3 ways to choose the first box and we want to choose any 4 of the 6 b***s to put into this = 3C(6,4) = 45 ways

And we want to put the other two b***s into either of the remaining two boxes = 2 ways to do this

The last box is empty by default

So 45 x 2 = 90 ways in this arrangement

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And we can put 3 b***s into one box and 3 into another

There are 3 ways to choose the first box and we want to choose 3 of the 6 b***s

So 3C(6,3) = 60 ways

And there are 2 ways to select the next box to put the remaining 3 b***s into = 2

The last box is empty by default

So 60 x 2 = 120 ways in this arrangement

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And finally, we can put 2 b***s into every box

We have 3 ways to choose the frst box and we want to choose 2 of the 6 b***s to put into that box  3C(6,2)  = 45 ways

And then we have 2 ways to select the next box and we want to choose 2 of the remaining 4 b***s to put into this  = 2C(4,2) = 12 ways

The last 2 b***s and box remain by default.

So 45 x 12 = 540 ways

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So, to recap

6 0 0 =  3 ways

5 1 0 = 36 ways

4 2 0 = 90 ways

4 1 1 = 180 ways

3 3 0 = 120 ways

3 2 1 = 360 ways

2 2 2 = 540 ways

So   3 + 36 + 90 + 180 + 120 + 360 + 540 = 1329 ways

I'll give this one a shot

I'm assuming that two b***s go into every box......(???)

We have two tasks.......choose a box.......choose the b***s to put in that box.

We have 3 ways to choose the first box and from the 6 b***s we want to choose 2 of them to put in this box.

Then, we have 2 ways to choose the second box and we want to choose 2 ballls from the remaining 4 to put in this box.

The third box and the last two b***s are already decided, by default.

So we have

3C(6,2)  x  2C(4,2)  = 540 ways

EDIT:  this might be the correct answer is the b***s and the  boxes were all INdistinguishable.

for that is the question that I attempted to answer.    LOL

I counted them and I get 7 ways

0,0,6

0,1,5

0,2,4

0,3,3

1,1,4

1,2,3

2,2,2

Since all the b***s and all the boxes are the same, the only difference can be is how many b***s  are in the boxes.

Well now, that is interesting.     LOL

I am not laughing at Chris, my answer could be wrong.  The 2 extreme answers are funny though.  :)

Melody has assumed the b***s and the boxes are both indistinguishable....

But.....the question clearly states that they are both distinguishable....capable of being specificallly identified....

Thus. all the b***s are different and all the boxes are different....

(???)

YES the question does clearly state that - next time I really must wear my glasses when i am reading questions.

I probably misread this same word in my other answers as well which would make them all wrong.  

The joke is on me     ROF LOL

Thanks Chris. 

CPhill's second answer may well be correct - that is certainly how I should have started it but I have not worked it out all through to the end. :)

Yeah...check my math, Melody....you're not the ONLY one wearing glasses...!!! LOL!!!

6 DIFFERENT B***S AND 3 DIFFERENT BOXES

THIS HAS BEEN EDITED

I BELIEVE CPHILL AND I ARE FINALLY IN TOTAL AGREEMENT

SO, THE PROBABILITY THAT IT IS THE CORRECT ANSWER HAS TO BE IMPROVED   (◐‿◑)

0,0,6        3*1 = 3            

0,1,5        3!*6= 36           

0,2,4        3!*6C2 = 6*15=90     

0,3,3       3!*6C3 = 6*20=120      

1,1,4       3!*6*5 = 180        

1,2,3       3!*6*5C2 = 6*6*10= 360   

2,2,2       3!*6C2*4C2 = 6*15*6=540   

$${\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{36}}{\mathtt{\,\small\textbf+\,}}{\mathtt{90}}{\mathtt{\,\small\textbf+\,}}{\mathtt{120}}{\mathtt{\,\small\textbf+\,}}{\mathtt{180}}{\mathtt{\,\small\textbf+\,}}{\mathtt{360}}{\mathtt{\,\small\textbf+\,}}{\mathtt{540}} = {\mathtt{1\,329}}$$

Which ones did we get different Chris?

I just checked - only 2 of the numbers are the same - all the others are different   LOL

Are you keen enough to check your or my answer again?  I am running out of steam and time for these questions.  

We have two tasks here....choose the boxes....choose the b***s to go into those boxes...

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4 2 0

We have 3  ways to choose the first box......and we want to put 4 of the 6 b***s into it

3C(6,4)  = 45

Then, out of the remaning 2 boxes, we want to choose one of the two to put the remaining two b***s into

So 45 x 2 = 90

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4 1 1

Agian we have 3 boxes to choose which 4 b***s out of 6 we put into it

So 3C(6,4)= 45 ways

And we have two ways to select the next box and two ways to select which ball goes into that box...

The last box and ball are defaulted

So we have

2C(2.,1) = 4 ways

So 45 x 4 = 180

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3 3 0

3 ways to choose the first box......and we want to choose 3 of 6 b***s to put into it

3C(6,3) = 60 ways

Then, we have 2 ways to choose the next box to place the remaining three b***s into

So 2 x 60  = 120 ways

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3 2 1

3 ways to choose the first box....choose 3 of 6 b***s to put into it   = 60 ways  

2 ways to select the next box and we want to choose 2 of the 3 remaining b***s to put into it

2C(3.2) = 6

Again, the last box and ball are defaulted

So   60 x 6 = 360 ways

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2 2 2

3 ways to select the first box .... choose 2 of 6 b***s to go into it  = 3C(6,2) = 45

2 ways to pick the next box and choose 2 of the remaining 4 b***s to put into it

2C(4,2)  = 12

The last box and remaig two b***s are defaulted

So 45 x 12 = 540 ways

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Anyway, that's what I get......!!!

This doesn’t give any restrictions other than both the b***s and boxes are distinguishable.

Here are solutions for certain restrictions or freedoms.

If any box(N) can contain zero to (k) b***s then the solution is

$$\displaystyle N^k \ = \ {3^6} \ = \ 729$$

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If a box MUST contain (1) but can contain up to (k-N+1) b***s then the solution is

$$\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{N-1}} *(-1)^i*\binom{N}{i}* (N-i)^k \\$$

This is a scripted link to Wolfram alpha

https://www.wolframalpha.com/input/?i=sum+%28+binom%283%2Ci%29*%28-1%29^i*%283-i%29^6%29+from+i%3D0+to+2

This is the script:

sum (binom(3,i)*(-1)^i*(3-i)^6) from i=0 to 2

(Note the value of 540 is consistent with CPhill’s solutions)

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If the boxes are distinguishable and the b***s are indistinguishable then

$$\displaystyle \left( {\begin{array}{*{20}c} 56 \end{array}} \right) = \dfrac{{(N+k-1)!}}{{k!}}\; \hspace{15pt}| \hspace{15pt} \Text {N=6 \; k=3} \\$$

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When both the b***s and boxes are indistinguishable then k parts of N partitions may be needed. http://mathworld.wolfram.com/PartitionFunctionP.html This might seems easier, but this is intermediate level combinatorics. The partition function is not any more difficult to understand than combinations, but the assembly of what to add, subtract and when, increases the complexity from b***h to royal b***h.

Good thing Melody appointed me Royal SOB.

$$\tiny \textcolor[rgb]{1,,0}{\text {(Edited: Formating )}}\\$$

Thanks....Nauseated.....I'll have to plow through that....

Now......

What if N distinguishable b***s were rolled along all the possible "right and up" paths of a 6 x 6 grid  ???

.....eh, never mind....

 We'll let some other SOB figure that one out......

Hi Nauseated,

If all 6 b***s are different and all 3 boxes are different   3^6=729  certainly sounds correct.

this is the answer that I had given for this scenario

http://web2.0calc.com/questions/how-many-ways-are-there-to-put-6-b***s-in-3-boxes-if-the-b***s-are-not-distinguishable-and-neither-are-the-boxes#r2

Added:   I appologize Nauseated - I did have that answer in that place but I had a different answer here.  I am just trying to sort the discrepancy out.  I can see hat 729 is correct but I cannot as yet see why my other answer is wrong.

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The rest of it I have no idea what you have found.  

I know you have found something but I don't understand what it is.

540 is only the number of ways you can put 2 b***s in each of 3 boxes when all the b***s and all the boxes are different from each other.

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Just come in on this conversation, here's my take on it.

First method: First ball can go into any of the three boxes, second ball can go into any of the three boxes etc, total number of arrangements is 3*3*3*3*3*3 = 729.

Second method: All six b***s could go into a single box, or they could be split five in one box and one in another, or they could be split 4 - 2 - 0, or 4 -1 -1 and so on, seven possible splits.

The number of possibilities for each split as follows.

6 - 0 - 0: 3 ways of doing this, (the 6 can be in any one of the three boxes).

5 - 1 - 0: 36 ways. There are 6 possible orders for the 5-1-0, and 6 possible choices for the odd ball.

4 - 2 - 0: 90 ways. Again 6 possible orders and 6C2=15 choices for the 2.

4 - 1 - 1: 90 ways. 6 possible orders, (3 possibilities for the 4, but double this because the two 1's are different), then 6C4=15 choices for the 4.

3 - 3 - 0: 60 ways. 3 possible orders (positions for the 0), and 6C3=20 choices for either 3.

3 - 2 - 1: 360 ways. 6 possible orders, 6 choices for the 1 and 5C2=10 coices for the 2 or 3.

2 - 2 - 2: 90 ways. 6C2=15 choices for the first 2, 4C2=6 choices for the second 2.

Total = 3 + 36 + 90 + 90 + 60 + 360 + 90 = 729. 

Yep I think we are all in agreement on this one.   We all keep good company :))

LOL  I just realised that I have answered this same question in two different places and got 2 different answers.

Well this was definitely one of my answers and it does make perfect sense.

I wonder if my other answer makes perfect sense as well ? 

There is always lots of humour to be had with probability and the various topics that go with it.  :)))

Bertie, I am hoping you can clear up some confusion for me please.

Obviously is there are 6 different b***s and 3 different boxes and any number of b***s can be put into any box then the number of possibilities is  3^6 = 729

Looking at all the different possibilities, yours must be correct because mine (and Chris's) dont add up to 729.

these are the ones that are not correct

1, 1, 4        put the boxes in a line.  there are 6 possible b***s that can be put in the first one, then 5 b***s that can be put into the seconds one and the remaining b***s go in  the last one. that is 6*5=30

Now that line could have been formed in 3!=6 ways.  So why isn't it 30*6=180 ways?      (you got 90)

Similar confusion comes up for 3,3,0 and 2,2,2. 

It has got to be something to do with more than one box having the same number of b***s but I don't get it.

This looks like an interesting conversation.

But i HAVE no idea what you're talking about.

Wish i could join in.

Hi Melody

The reason that you get the wrong answer for the 1-1-4 arrangement is that you count some of the possibilities twice.

Suppose that the two odd b***s (if I can call them that !) are labelled A and B, so that there are those two as singles and the others as a group of 4.

They can be placed in the three boxes in 3! = 6 different ways. A-B-4, B-A-4, A-4-B, B-4-A, 4-A-B and 4-B-A.

Now, how many A, B, 4 combinations are there ? The answer is 6C4 = 15, the number of ways in which the group of 4 can be taken from the original 6, (the two singles take care of themselves), not 6*5 = 30.

The problem with the 6*5=30, is that it takes order into account, that is, it sees A,B,4 and B,A,4 as being different and as such covers two of the 6 possibles listed earlier rather than just one.

To look at the 3-3-0 arrangement, suppose that a group of three is chosen and called 3(1), and that the remaining three are called 3(2).

Now suppose that they occupy the first two boxes, so that we have 3(1)-3(2)-0 or 3(2)-3(1)-0. How many possible combinations are there there ? It's tempting to say 6C3 for the first and 6C3 for the second making 2*6C3 = 2*20 = 40 in all. (That, in effect, is what is being done if you say that there are 3!=6 different 3,3,0 arrangements.) That's wrong though, it's actually just 6C3=20, because 6C3 includes both 3(1) and also 3(2).

The 0 can occupy three different positions, so there are 3*20 = 60 combinations in all.

It's easy if you look at it as

box 1    box 2    box 3

3            3           0

3            0           3

0            3           3

There are 6C3 possibilites for each row making 3*6C3 = 60 possibles in all.

The 3!=6, in effect, counts each row twice.

Thanks Bertie, I think I understand perfectly now.  Hopefully I will get it right next time. 

No I really do think I have it.

I would do the second one a little differently.  I'd stick to the 6 possible permutations but I would halve the number of combinations.  It would be easy to miss though. :/

I haven't tried the 2,2,2, on yet - that will be the test LOL  

♬♪♫ ヾ(*・。・)ﾉ ♬♪♫

I got it. I think I really have got it. 

I  hope these probability/combination/permutation questions keep coming in, I want more practice .    

Thanks again Bertie :)

We have 3 choices for each ball, and each choice is independent. So 3^6 = 729 ways to place the b***s.

If all are distguinshable then answer is 3^6=729