q43

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Saumyojit

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Six white and six black balls of the same size are distributed among ten urns so that there is at least one ball in each urn. What is the number of different distributions of the balls?​


Case 1: (6 w , 4 bl) -->Assume 6 white balls are taking the first six urns (urn 1-6) , 4 black balls are taking the next four Urns (urn 7-10), out of 2 same black balls remaining each has 10 urns Options to go in .


One eg : w w w w w w b b b b (Ten urns filled ) Now two white balls from urn 5 and urn 6 if they go to urn 7 & 8 (4c2 ) , then there is another new arrangement can be w w w w b b w w b b .....SO

4c1 + 4c2 + 4c3 + 4c4 - > these are all counting when white balls are shifting from their original position to some other Urns .

No of ways : (4c1 + 4c2 + 4c3 + 4c4) * (10 * 10 ) /2 = .. ...

Case 2 : (4w , 6 bl) --> No of ways : (4c1 + 4c2 + 4c3 + 4c4) * (10 * 10 ) /2 = .. ...


Case 3 : 5w 5bl --> (5c1 + 5c2 + 5c3 + 5c4) * (10 * 10 ) = ...


Answer is not coming right.
 
are URNS identical ? Answer is : 26,250

Another approach I came up with
6white , 4 black balls--> 10c6 * 4c4 * (10*10)/2= ...

6 black ,4 white = 10c6*4c4 *( 10*10)/2

5 bl, 5 w= 10c5 *5c5 * 100= 25200

Adding them up gives 46,200

Something is wrong



So if they are identical then what is the meaning of Distribution of balls ?
W,(w,b) w w (w,b) w b b b b --> this is one distribution and this is
W,(w,b) b b (w,b) w w w bb Another ?
 
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Saumyojit said:
are URNS identical ? Answer is : 26,250

Another approach I came up with
6white , 4 black balls--> 10c6 * 4c4 * (10*10)/2= ...

6 black ,4 white = 10c6*4c4 *( 10*10)/2

5 bl, 5 w= 10c5 *5c5 * 100= 25200

Adding them up gives 46,200
Click to expand...
Please explain why, when the question says 6 black, 6 white, you are using other numbers. You aren't explaining your thinking enough to be sure whether it is right.

My first thought would be to take cases according to which urn(s) get "extra" balls (3 in one urn, or 2 in one, 2 in another), and then break down further by the colors in each (e.g. BBB, BBW, BWW, WWW for the first case). I suspect you are double-counting by placing 10 balls first and then placing the last two but not considering that you can get the same result multiple ways.
 
Dr.Peterson said:
Please explain why, when the question says 6 black, 6 white, you are using other numbers. You aren't explaining your thinking enough to be sure whether it is right.

My first thought would be to take cases according to which urn(s) get "extra" balls (3 in one urn, or 2 in one, 2 in another), and then break down further by the colors in each (e.g. BBB, BBW, BWW, WWW for the first case). I suspect you are double-counting by placing 10 balls first and then placing the last two but not considering that you can get the same result multiple ways.
Click to expand...


Case 1: (6 w , 4 bl) -->

Any six out of 10 urns where white will be placed and in the remaining 4 , 4 black will be placed .

10c6 * 4c4 * (10*10)/2
Filling up 10 urns first then filling any two urns with any two identical balls-->10*10

urn 1- black ball
urn 2 -black ball

urn 2- black ball
urn 1 -black ball

But vice versa case is getting counted that's why
100/2


Just say that Is this case of 6w ,4bl is right?


Dr.Peterson said:
you are using other numbers.
Click to expand...

What other numbers ?
 
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Saumyojit said:
What other numbers ?
Click to expand...
You used 6 and 4, not 6 and 6. What else could I mean??
Saumyojit said:
Case 1: (6 w , 4 bl) -->

Any six out of 10 urns where white will be placed and in the remaining 4 , 4 black will be placed .

10c6 * 4c4 * (10*10)/2
Filling up 10 urns first then filling any two urns with any two identical balls-->10*10

urn 1- black ball
urn 2 -black ball

urn 2- black ball
urn 1 -black ball

But vice versa case is getting counted that's why
100/2


Just say that Is this case of 6w ,4bl is right?
Click to expand...
So you're placing 6 white and 4 black, and THEN placing the other two black balls? You didn't say that. Communication is an important part of math, and I am trying to encourage you to do so. As I've said before, if you don't write what you mean so that others can understand, you may not know what you mean yourself.

Do you not see the double-counting here? I avoid schemes like this where I place some, and then others, because the "others" might have been the "some", yielding the same result. For instance, WB WB W W W W B B B B, where I added the B's in the first two urns, is the same arrangement as BW BW W W W W B B B B, where I first placed 6 black and 4 white, and then added the W's in the first two urns.

I also have to say that it is very difficult to judge someone else's scheme; trying to understand your thinking makes me dizzy. It's hard enough to follow my own thinking in problems like this!

I suggest you try solving a similar problem with smaller numbers such that you can actually list the possibilities and test your thinking. Try this, for example (I haven't yet):

2 white and 2 black balls, otherwise indistinguishable, are distributed among 6 urns so that there is at least one ball in each urn. What is the number of different distributions of the balls?​
 
Dr.Peterson said:
WB WB W W W W B B B B, where I added the B's in the first two urns, is the same arrangement as BW BW W W W W B B B B, where I first placed 6 black and 4 white, and then added the W's in the first two urns.
Click to expand...
yes that's why I said consider only the case 1 . We are double counting when we are considering both these cases (6w,4bl) and (4w ,6bl ) . But i said only the first case .

(6w,4bl) -->10c6 * 4c4 * (10*10)/2 For this case this is right . SO this expression is also covering arrangements of 4w,6bl and 5w,5bl . Isn't it?


Dr.Peterson said:
I suggest you try solving a similar problem with smaller numbers
Click to expand...
How will i fill 6 urns with 4 balls . It has to be more . Please solve the OP only .
 
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Saumyojit said:
yes that's why I said consider only the case 1 . We are double counting when we are considering both these cases (6w,4bl) and (4w ,6bl ) . But i said only the first case .
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If you are using cases, then you must define each case clearly so that they are mutually exclusive. You have not done so, as far as I can see. So it is useless to count one case.
Saumyojit said:
How will i fill 6 urns with 4 balls . It has to be more . Please solve the OP only .
Click to expand...
I chose the wrong numbers; you can choose numbers that make sense. My strategy is still a good one.

I am not here to solve the problem. I have no particular interest in it. I am here to help you think, and you are refusing. So I'll go back to ignoring the thread.
 
Let’s form a decision tree: first you would decide to fill the 10 urns with 10 balls, and there are three ways you can decide to fill: 1) 6w4b 2) 6b4w and 3) 5b5w. Then you would make another decision where you put the extra 2 balls. Am I correct so far?
 
I think I have made the same approach mistake in the OP. Where you put the 1 ball in 1 urn doesn’t matter much in this problem. It all starts with deciding which urn to put in more than 1 ball. There are two cases which one can make such decision. Case 1: put 3 balls in 1 urn and just fill in the rest one by one. Case 2: put 2 balls in 1 urn and 2 in another and fill in the rest one by one.

CASE1
When deciding which urn to put in 3 balls, the 3 balls can have the following pattern: 3b, 3w, bbw, bww.

3b: choose 1 urn: 10c1; choose 3 urns out of the 9 left and put in the 3 black balls: 9c3; put the white in the rest 6c6 = 10c1*9c3*6c6 = 840
3w: same result as 3b = 840
bbw: choose 1 urn: 10c1; choose 3 urns out of the 9 left and put in the 4 black balls: 9c4; put the white in the rest 5c5 = 1260
bww: same result as bbw = 1260

CASE2
When deciding 2 urns to put 2 2-balls, the 2 2-ball-combos can have the following pattern: 2b x2, 2w x2, bw x2, 2b+bw, 2w+bw and 2w+2b.

2b x2: choose 2 urns 10c2, choose 2 urns out of 8 to put the rest of the black balls: 8c2, and put the white in the rest: 6c6 = 10c2*8c2*6c6=45*28=1260
2w x2: same result as 2b x2=1260
bw x2: same process but instead of 8c2, use 8c4 = 3150

2b+bw: choose 2 urns but this time the sequence of your choice matter, so 10p2, then choose 3 urns out of 8 to put the rest of the black balls: 8c3, and put the white in the rest, and 10p2*8c3*5c5 = 5040
2w+bw: same result as 2b+bw = 5040
2w+2b same process but instead of 8c3 use 8c4 = 6300

And add them altogether: 840+840+1260+1260+1260+1260+3150+5040+5040+6300 = 26250
 
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Dr.Peterson said:
If you are using cases, then you must define each case clearly so that they are mutually exclusive.
Click to expand...
ok.

Just consider these two cases
Case 1: (6w,4bl) -->10c6 * 4c4 * (10*10)/2
placing 6 white and 4 black, and THEN placing the other two black balls in any one urn or two diff urns .

Case 2: (6bl,4w) -->10c6 * 4c4 * (10*10)/2
placing 6 black and 4 w and THEN placing the other two black balls.

Now i feel that I Am already counted SOME of the arrangements of case 2 in case 1 .

Some eg's i worked out were :
WB WB W W W W B B B B is the same arrangement as BW BW W W W W B B B B

W W WB W W B B WB B B is same as W W BW W W B B BW B B

(5bl 5w) W W WB W W BW B B B B is same as W W WB W W WB B B B B (6w, 4bl)

Below
These are Unique arrangements .
6w,4bl: W WBB W W B B B W W B

6bl,4w: B BWW B B B B W W W W

5bl 5w: BWB B B B B W W W W W

5bl 5w: BW BB B B B W W W W W


Shall set inclu-exclusion will be used here?
 
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Saumyojit said:
Just consider these two cases
Case 1: (6w,4bl) -->10c6 * 4c4 * (10*10)/2
placing 6 white and 4 black, and THEN placing the other two black balls in any one urn or two diff urns .

Case 2: (6bl,4w) -->10c6 * 4c4 * (10*10)/2
placing 6 black and 4 w and THEN placing the other two black balls.

Now i feel that I Am already counted SOME of the arrangements of case 2 in case 1 .
Click to expand...
Since these cases are not mutually exclusive, they really are not cases at all. When you divide a problem into cases, they must be distinct outcomes, not just different procedures to place things.

Maybe you could rescue your method, but I don't think it is appropriate, and am not going to try to help you use a bad method.
 
Dr.Peterson said:
you could rescue your method
Click to expand...
@sky2rain @pka @Jomo


Case 1: (6w,4bl) -->10c6 * 4c4 * (10*10)/2
placing 6 white and 4 black, and THEN placing the other two black balls in any one urn or two diff urns .

Case 2: (6bl,4w) -->10c6 * 4c4 * (10*10)/2
placing 6 black and 4 w and THEN placing the other two black balls.

Case 3 : (5bl,5w) --> 10c5 * 5c5 * (10*10)


Some eg's i worked out were :
WB WB W W W W B B B B is the same arrangement as BW BW W W W W B B B B

W W WB W W B B WB B B is same as W W BW W W B B BW B B

(5bl 5w) W W WB W W BW B B B B is same as W W WB W W WB B B B B (6w, 4bl)

We have to exclude arrangements of these 3 types


Summing up after excluding all the double counting i think will give the correct answer .

But how to exludce the double counting? Nothing is coming to my mind
 
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Saumyojit said:
@sky2rain @pka @Jomo


Case 1: (6w,4bl) -->10c6 * 4c4 * (10*10)/2
placing 6 white and 4 black, and THEN placing the other two black balls in any one urn or two diff urns .

Case 2: (6bl,4w) -->10c6 * 4c4 * (10*10)/2
placing 6 black and 4 w and THEN placing the other two black balls.

Case 3 : (5bl,5w) --> 10c5 * 5c5 * (10*10)


Some eg's i worked out were :
WB WB W W W W B B B B is the same arrangement as BW BW W W W W B B B B

W W WB W W B B WB B B is same as W W BW W W B B BW B B

(5bl 5w) W W WB W W BW B B B B is same as W W WB W W WB B B B B (6w, 4bl)

We have to exclude arrangements of these 3 types


Summing up after excluding all the double counting i think will give the correct answer .

But how to exludce the double counting? Nothing is coming to my mind
Click to expand...
I wanna point out one thing. Instead of excluding, think of the factual matter. The 10*10 after 10c6 in your formula, I would guess you are choosing any of 10 urns after 10c6? The factual matter is when you add something after you arranged 10 items in urns, it’s equivalent of emptying an urn out and refill something different. However If you’re emptying one, the 10c6 would no longer stand. It should be 9c something. So kept thinking there are always 10 urns to place items is already making the solution flawed. Does my solution make sense to you? I already have 26250 in #13.
 
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Saumyojit said:
I dont understand how am i emptying anything .

WWWWWW BBBB - > 10c6 * 4c4

W WB W WB W W B B B B -> 10 * 10

W WBB W W W W B B B B --> 10 *10


What is wrong?
Click to expand...
I think you are well aware that in the case if you initially arranged 6 Bs out of 10 in this way:
W B W B W W B B B B, and again urn #2 and #4 are chosen to put two Ws, the double counting occurs.

Decision 1: you chose 6 urns out of 10 for W. Decision 2: you chose 2 urns out of 10 for WB. Apparently after the second decision, you need to subtract 2 from your first decision to make the 2 Ws go away. And it’s not 10*10 as well. It’s 10c2 because the newly formed WBs cannot distinguish one from another, hence choosing urn #4 for 1st WB and #2 for second is the same as #2 for the 1st and #4 for the second. Once you have 10c2, you subtract the 2ws from your first decision to make it 8c4. And Bs don’t change. It then becomes exactly the same as my formula: 10c2*8c4*4c4
 
Hence instead of thinking adding two Bs to two urns, think of it this way: you chose two urns that initially contained Ws, emptied those two urns, and refilled them with WBs.
 
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