If the problem changes to: ten distinct transparent test tubes with diameter slightly larger than a ball, and 6 white and 6 black balls must go in them and at least 1 per tube, how many ways would the tubes appear? In this one your method would probably work. In our problem of urns, when you change your decision sequence, such as put in 10 first and 2 later, compare to 2 then 2 then 8 later or any sequence you prefer, it doesn’t change the consequence at all.
q43
- Thread starter Saumyojit
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@sky2rain @Dr.Peterson
I am getting confused.
(6b, 4w) = W BW W BW W W B B B B
(4b, 6W)= W WB W WB W W B B B B
Are you saying Double counting occurs only in this case (6b,4w) itself or when I am counting the second case (4b,6w) , then yes (4b,6w) is counting some arrangements which are already present in 1st Case.
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=
I am excluding the double arrangements .
Case 1 -Case 2 similarities = in case 1 Two black in two diff white urns and in case 2 Two white in two diff black urns . (W WB W WB W W B B B B = W BW W BW W W B B B B )
CAse 1 - Case 3 =Case 1 -Two black in a white( W WBB W W W W B B B B) & Case 3- One white and one black in a Black (W BWB W W W W B B B B )
Another similarity = Case 1: Two black one in white urn and another in black urn (W W W W W WB BB B B B ), Case 3: One white in black urn & one black in another black urn ( W W W W W BW BB B B B)
Another similarity = Case 1 : Two black in two white (W WB WB W W W B B B B ) and Case 3: One white in black and one Black in White (W WB BW W W W B B B B)
Case 2 - Case 3 : One white in white urn and one white in black urn and case 3 = one white in white and one black in white (WW WB W W W B B B B B)
Another similarity of Case 2-3 = Case 2: Two white in black and Case 3: One white and one black in white
ANother of case 2- case 3 = One white in black URN and one white in Another black urn & & Case 3 = One black in white urn and one white in black urn .
THese are the double countings coming . Am i mssing Some double countings then point out PLEASE .
6 black ,4 white = 10c6*4c4 *( 10*10)/2
5 bl, 5 w= 10c5 *5c5 * 100=
I am excluding the double arrangements .
Case 1 -Case 2 similarities = in case 1 Two black in two diff white urns and in case 2 Two white in two diff black urns . (W WB W WB W W B B B B = W BW W BW W W B B B B )
CAse 1 - Case 3 =Case 1 -Two black in a white( W WBB W W W W B B B B) & Case 3- One white and one black in a Black (W BWB W W W W B B B B )
Another similarity = Case 1: Two black one in white urn and another in black urn (W W W W W WB BB B B B ), Case 3: One white in black urn & one black in another black urn ( W W W W W BW BB B B B)
Another similarity = Case 1 : Two black in two white (W WB WB W W W B B B B ) and Case 3: One white in black and one Black in White (W WB BW W W W B B B B)
Case 2 - Case 3 : One white in white urn and one white in black urn and case 3 = one white in white and one black in white (WW WB W W W B B B B B)
Another similarity of Case 2-3 = Case 2: Two white in black and Case 3: One white and one black in white
ANother of case 2- case 3 = One white in black URN and one white in Another black urn & & Case 3 = One black in white urn and one white in black urn .
THese are the double countings coming . Am i mssing Some double countings then point out PLEASE .
Cross-cases. And think about your “cases”. What common and distinctive traits do they present to you. The least amount of one ball urns you need is 8, and the most is 9. So what you decide with 8 urns and 9 urns will force you to decide what happens with the all 10. Now think reversely. What you decide with 1 or 2 urns which you would put more than 1 ball will force you to decide how to distribute in 10 urns. If the question asks you to put in 10 balls in each urn with 1 ball at first and write a number on the last ball(s) and put the balls in urn according to the number written, then you will do 10c6*10*10, because in the example you present, the #2 and #4 will be written on the white balls in your (6b 4w) case, and on the black balls in your (6w 4b) case. This extra step adds distinguishing feature to your decision making and makes your instances mutually exclusive. Without this, the arrangements can’t be distinguished with what you decide to fill 10 urns first then 1 or 2 urns later.
Since you are so stubborn on the double counting, I will analyze this one for you: W WBB W W W W B B B B versus W BWB W W W W B B B B. Put it this way, the number 2 urn, let’s make it undecided and you have 2 choices to make, W, B, then after that decision you dump last 2 balls in it. It’s exactly the same as deciding to have 6w+4b or 5w+5b then dump the last 2 balls in it. This is just saying when you have 5w + 4b + 1 urn undecided, it’s possible for it to become 6w+4b or 5w+5b. So 5w + 4b + 1urn undecided=2*10*9c5. (2 basically tells you you have two choice of color)
So to reverse this process the double counting is so obvious! In your 6w+4b and 5w+5b cases, let’s just make 1 urn undecided. 6 urns contain w in 6w+4b can be reversed to 1 undecided to have 5w+4b+1urn undecided, which is 6*10c6, and 5 urns contain w in 5w+4w can be reversed to 1 undecided to have 5w+4b+1urn undecided, which is 5*10c5. If the forward process and reverse process are exactly the same, 6*10c6+5*10c5 should equal to 2*10*9c5. And guess what, they both equal to 2520!
If the problem asks you to put 5w and 4b balls each one in 9 urns first, then decide a color and mark it to put in the last urn, it is exactly the same as your 6w+4b or 5w+5b cases. And after putting in the last urn with marked ball your formula: 10c6*10 and 10c5*10 etc etc would work!
However the problem didn’t tell you to mark the last ball you decide. 5w + 4b + 1urn undecided=2*10*9c5 didn’t happen. The problem basically tells you to dump all three balls in that undecided urn, so you can’t have the 2 as a multiplier. it’s 1*10*9c5.
So to exclude double counting: (6*10c6+5*10c5)/2. (Check it yourself if that = 10*9c5)
And among your 6b+4w and 5w+5b cases the same thing happened. Use (6*10c6+5*10c5)/2 again and this reflects exclusions of double counted instances by growing 5b+4w into 6b+4w or 5w+5b.
So if this is so obvious that you grew double counted instances from 9c5, why can’t you just use 9c5?
So to reverse this process the double counting is so obvious! In your 6w+4b and 5w+5b cases, let’s just make 1 urn undecided. 6 urns contain w in 6w+4b can be reversed to 1 undecided to have 5w+4b+1urn undecided, which is 6*10c6, and 5 urns contain w in 5w+4w can be reversed to 1 undecided to have 5w+4b+1urn undecided, which is 5*10c5. If the forward process and reverse process are exactly the same, 6*10c6+5*10c5 should equal to 2*10*9c5. And guess what, they both equal to 2520!
If the problem asks you to put 5w and 4b balls each one in 9 urns first, then decide a color and mark it to put in the last urn, it is exactly the same as your 6w+4b or 5w+5b cases. And after putting in the last urn with marked ball your formula: 10c6*10 and 10c5*10 etc etc would work!
However the problem didn’t tell you to mark the last ball you decide. 5w + 4b + 1urn undecided=2*10*9c5 didn’t happen. The problem basically tells you to dump all three balls in that undecided urn, so you can’t have the 2 as a multiplier. it’s 1*10*9c5.
So to exclude double counting: (6*10c6+5*10c5)/2. (Check it yourself if that = 10*9c5)
And among your 6b+4w and 5w+5b cases the same thing happened. Use (6*10c6+5*10c5)/2 again and this reflects exclusions of double counted instances by growing 5b+4w into 6b+4w or 5w+5b.
So if this is so obvious that you grew double counted instances from 9c5, why can’t you just use 9c5?
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6 urns contain w in 6w+4b can be reversed to 1 undecided to have 5w+4b+1urn undecided, which is 6*10c6, and 5 urns contain w in 5w+4w can be reversed to 1 undecided to have 5w+4b+1urn undecided <- in this sentence there are 2 typos. 5w + 4w should be 5w+5b, and 5 urns contain w should be 5 urns contain b.
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So you are saying this approach is wrong even i exclude duplicate arrangements
I am showing you where the duplicates come from. It basically came from decide what to put to the next urn.
If you fixates on an approach of filling 1 ball into every urn of the 10 urns, then I could argue that it’s the same as you picked each urn and made a decision to fill 1 of either color. And this lays the logic that the double count can happen any time during the process. So take this example: if you have 2w and 1b and 1 undecided urn to fill either w or b, you could end up with 3w1b or 2w2b, and any of the 3w could be the urn which you have made the choice, so 3*4c3 and for 2w2b that’s 2*4c2 and that is 12+12=24.
With the 2w1b +1 undecided, you do 2*4*3c2=24, the 2 multiplier being the choices of colors you have. Both processes give you the same result, because you are filling 4 urns with b or w color basing on how you filled 3 urns and what you decide for the 4th.
If I tell you that for the undecided you can only choose red, you can still swap 1 w to 1 red for the supposedly chosen 2w2b and 3w1b, they will just turn into many 2w+1r+2b and 2w+1r+1b, 24 of them. But 12 of them will be double counted since when I force you to choose red for the undecided, the 2w1b+1undecided formula become 1*4*3c2=12 because the undecided spot has turned to decided.
the 9 urns with 1 ball and 1 urn with 3 balls case in our original problem, whatever was previously chosen for that 1 urn, w or b doesn’t matter anymore. After you finished, if the other 9 urns’ colors are known to you, that 1 urn with 3 balls has a fixed color combo, in the sense that it has been force decided. And no matter how you rearrange the other balls in 9 urns, that combo stays the same.
If you fixates on an approach of filling 1 ball into every urn of the 10 urns, then I could argue that it’s the same as you picked each urn and made a decision to fill 1 of either color. And this lays the logic that the double count can happen any time during the process. So take this example: if you have 2w and 1b and 1 undecided urn to fill either w or b, you could end up with 3w1b or 2w2b, and any of the 3w could be the urn which you have made the choice, so 3*4c3 and for 2w2b that’s 2*4c2 and that is 12+12=24.
With the 2w1b +1 undecided, you do 2*4*3c2=24, the 2 multiplier being the choices of colors you have. Both processes give you the same result, because you are filling 4 urns with b or w color basing on how you filled 3 urns and what you decide for the 4th.
If I tell you that for the undecided you can only choose red, you can still swap 1 w to 1 red for the supposedly chosen 2w2b and 3w1b, they will just turn into many 2w+1r+2b and 2w+1r+1b, 24 of them. But 12 of them will be double counted since when I force you to choose red for the undecided, the 2w1b+1undecided formula become 1*4*3c2=12 because the undecided spot has turned to decided.
the 9 urns with 1 ball and 1 urn with 3 balls case in our original problem, whatever was previously chosen for that 1 urn, w or b doesn’t matter anymore. After you finished, if the other 9 urns’ colors are known to you, that 1 urn with 3 balls has a fixed color combo, in the sense that it has been force decided. And no matter how you rearrange the other balls in 9 urns, that combo stays the same.
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You got to know where they come from in order to exclude it, but if you know where they come from, the previous 8 or 9 choices have forced you with the last decision(s) already so why not just approach based on 8 urns and 9 urns of choices instead of 10? you don’t need to exclude anything!
@sky2rain
I am really determined to understand your opinion.
I am stuck in this paragraph . You say there is one urn and 2white ball & 1 black ball and i can fill any one colour ball right?
Either 2 whites or 1 black . How will i end up with 3 white 1 black or 2 white 2black. I don't understand . Only explain this paragraph . Lets proceed concisely step by step not all in one go .
Another typo, 2 places for b in 2w2b that undecided can become a b. Just list the 12 possible situations for 2w1b1u for w=white b=black and u= undecided. You will have 24 permutation sets if u turns into either w or b. Hence it would be the same 24 ways to change back to what was supposedly to be a u (equivalent to emptying an urn). And then you make that u to be a red, yellow, 2b +1w in that one urn (In this simple example if assumed you have total of 3 black balls and 3 white balls, so the last balls left are: 2 blacks for 3w1b and 1 white 1 black for 2w2b) or a teddybear, they all give you 12 unique instances, but you counted 24 across both cases for 3w1b and 2w2b.
And I found a simpler way explaining it because you had trouble visualizing “emptying an urn.” Just for 4 urns containing a white or black ball in every urn, in the cases that formed arrangements of 3w1b and 2w2b, if you are told to pour cement into in any of the 3 urns containing w in 3w1b, and in any of the 2 containing b in 2w2b, and after the pouring the balls used to be in the urn become unrecognized, there are 24 ways to do it but there are only 12 total unique outcomes. (P.S. part is a detailed description of why the original problem is like pouring cement, just for the sake of thinking process)
However the problem didn’t ask you how many ways are there to do it, but how many kinds of arrangement that would be eventually presented to us.
P.S.
For the original problem, when putting the last 2 balls left in your hand to any urn containing w in 3w1b, or to any urn contain b in 2w2b, it’s exactly the same as pouring cement into the urn to make what used to be in that urn unrecognized and the eventual outcomes become the same.
When you are left with 2 black balls in your hand while already arranged 3w1b into 4 urns with each containing 1 ball, or when you are left 1 white 1 black balls while already arranged 2w2b into 4 urns each containing 1 ball, just by dumping 2 black balls into 1 urn containing w in the 3w1b case, compare to dumping 1 white and 1 black ball into 1 urn containing b in the 2w2b case, they both make what used to be the urn as a w or b unrecognized because the eventual outcome is that 1 urn will contain 2 blacks and 1 white, regardless what the urn used to contain.
So the double counting comes from cases that used to be different, say wbww vs wbwb, it differs on the last urn. And by filling that urn both with cement to make what used to be recognizable as different to the same as wbwc and wbwc, there a double counting occurs.
However the problem didn’t ask you how many ways are there to do it, but how many kinds of arrangement that would be eventually presented to us.
P.S.
For the original problem, when putting the last 2 balls left in your hand to any urn containing w in 3w1b, or to any urn contain b in 2w2b, it’s exactly the same as pouring cement into the urn to make what used to be in that urn unrecognized and the eventual outcomes become the same.
When you are left with 2 black balls in your hand while already arranged 3w1b into 4 urns with each containing 1 ball, or when you are left 1 white 1 black balls while already arranged 2w2b into 4 urns each containing 1 ball, just by dumping 2 black balls into 1 urn containing w in the 3w1b case, compare to dumping 1 white and 1 black ball into 1 urn containing b in the 2w2b case, they both make what used to be the urn as a w or b unrecognized because the eventual outcome is that 1 urn will contain 2 blacks and 1 white, regardless what the urn used to contain.
So the double counting comes from cases that used to be different, say wbww vs wbwb, it differs on the last urn. And by filling that urn both with cement to make what used to be recognizable as different to the same as wbwc and wbwc, there a double counting occurs.
Dr.Peterson
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As I've said, I have no interest in helping you fix up a complicated method. But in case it matters, let me show you the method I just used to solve the problem in a few minutes (with a few errors along the way):
There are either one or two urns with more than one ball.
Case 1: 1 urn has 3 balls, 9 urns have 1 each
Case 1a: BBB in the one urn, 3B6W arranged in others; 10 ways to chose the one, C(9,3) ways to place the other B's.
Case 1b: BBW in the one urn, 4B5W arranged in others; 10 ways to chose the one, C(9,4) ways to place the other B's.
Case 1c: BWW in the one urn, 5B4W arranged in others; 10 ways to chose the one, C(9,5) ways to place the other B's.
Case 1d: WWW in the one urn, 6B3W arranged in others; 10 ways to chose the one, C(9,6) ways to place the other B's.
(1c = 1b, 1d = 1a)
Total: 4200
Case 2: 2 urns have 2 balls each, 8 urns have 1 each
Case 2a: BB,BB in 2 urns, 2B6W arranged in others; C(10,2) ways to choose the two, C(8,2) ways to place the other B's.
Case 2b: BB,BW in 2 urns, 3B5W arranged in others; P(10,2) ways to choose the two (distinct), C(8,3) ways to place the other B's.
Case 2c: BB,WW in 2 urns, 4B4W arranged in others; P(10,2) ways to choose the two (distinct), C(8,4) ways to place the other B's.
Case 2d: BW,BW in 2 urns, 4B4W arranged in others; C(10,2) ways to choose the two, C(8,4) ways to place the other B's.
Case 2e: BW,WW in 2 urns, 5B3W arranged in others; P(10,2) ways to choose the two (distinct), C(8,5) ways to place the other B's.
Case 2f: WW,WW in 2 urns, 6B2W arranged in others; C(10,2) ways to choose the two, C(8,6) ways to place the other B's.
(2e = 2b, 2f = 2a)
Total: 22,050
Grand total: 26,250
There is probably a nicer way, but at least this is an orderly set of cases.
So to sum this up, basically you counted the number of ways to go through the process (misled by your visualization of the problem) and you would be correct if there exists a weird question asking: how many ways are there to fill 10 urns, 1 ball in each urn first and put in the last two balls in any 1 or 2 urns, if you have 6 black balls and 6 white balls? But our problem asks for how many arrangements will it present after any process of filling the urns, as long as there exists at least one ball in a urn.
I understood your approach . it seems that thinking and approach of mine is same . But as long nobody can puncture of where am i wrong in this below step , i cannot move on even if i am trying hard to move on as its very imp to know where is my approach wrong.
@Dr.Peterson
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=
I am excluding the double arrangements .
Case 1 -Case 2 similarities = in case 1 Two black in two diff white urns and in case 2 Two white in two diff black urns . (W WB W WB W W B B B B = W BW W BW W W B B B B )
CAse 1 - Case 3 =Case 1 -Two black in a white( W WBB W W W W B B B B) & Case 3- One white and one black in a Black (W BWB W W W W B B B B )
Another similarity = Case 1: Two black one in white urn and another in black urn (W W W W W WB BB B B B ), Case 3: One white in black urn & one black in another black urn ( W W W W W BW BB B B B)
Another similarity = Case 1 : Two black in two white (W WB WB W W W B B B B ) and Case 3: One white in black and one Black in White (W WB BW W W W B B B B)
Case 2 - Case 3 : One white in white urn and one white in black urn and case 3 = one white in white and one black in white (WW WB W W W B B B B B)
Another similarity of Case 2-3 = Case 2: Two white in black and Case 3: One white and one black in white
Another of case 2- case 3 = One white in black URN and one white in Another black urn & & Case 3 = One black in white urn and one white in black urn .
These are the double countings coming .
Now excluding the cases of case1 from case 3 -->6c2 --> 15
Excluding from case 3 --> 5c1 * 5c1 = 25
Excluding From case 3 --> 5c1 * 4c1 =20
Excluding case 2 From case 1 --> 6c2 --> 15
Excluding case 3 From case 2 --> 5c1 * 4c1 --> 20
Excluding case 3 --> 5c1 =5
Excluding case 3 --> 5c1 * 5c1 =25
Total arrangements of case 3 --> 10c5 * 5c5 * 100 = 25200
25200 - ( 15 + 25 + 20 + 20 + 5 + 25) =25090
Case 2 = 10c6 * 4c4 * (10 * 10) /2 = 10,500
10,500 - 15= 10,485
Case 1 --> 10c6 * 4c4 * 50 = 10,500
Total arrangements excluding = 10,500 + 25090 + 10485 = 46075
I am counting more
Dr.Peterson
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No, your approach is not the same as mine. I am unwilling to try to help you with a method in which you have to exclude duplicates among cases; that is far too complicated, and unnecessary. When you use cases, they must be mutually exclusive. (Yes, inclusion-exclusion can be a valid method, but that is not really what you are doing.) Your approach is wrong simply in being too complicated to keep track of!
When you say "puncture", do you mean "point out"? That is not what it means, and I can't think of anything else you might intend.
If you insist on doing this please try with 3w 3b in 4 urns first. It’s essentially the same. That way you can draw all counts on a piece of paper and find which case has doubles on which visually
Just explain in 6w4b what is (10*10)/2. I can understand what you are trying to get while doing 10c6*4c4, but it seems to me that (10*10)/2 means you are aware the first 10 choices for filling in the 11th ball and latter 10 choices for filling in the 12th ball, they create a double and you have to divide it by 2. So clearly it implies that if you reverse your decision i.e. if the 11th fill as a black ball and 12th as a black ball, and filling in 12th first then 11th later, they wouldn’t make a difference. And you have to exclude by dividing by 2.
So if I am to interpret your formula as decisions to form arrangement in 6w4b, here are the decisions: Decision 1: pick 6w and 4b balls to put in 10 urns in some order. Decision 2: must put the 2 black balls in one or two urns (forced to be so as 6w6b-6w4b=2b). Decision 3: filling the first black ball and Decision 4: filling the 2nd black ball. This is gives a formula without exclude anything: 10c6*4c4*10c1*10c1. Decision 1 is reflected by 10c6*4*4c4, Decision 2 is reflected by c1, and Decision 3 and 4 basing on Decision 2 are 10c1 and 10c1.
Since Decision 3 and 4 can be made in reversed order as 10c1*10c1 = 10c1*10c1, and reduce two decision to a single decisions: ((10*10)/2)c1, and now it is exactly as yours: 10c6*4c4*((10*10)/2)c1. To avoid confusion we call this reduced decision Decision Three or Decision T = (10*10)/2. So your formula basically can be written as D1*(DT basing D2). And with the logic D3*D4 = D4*D3, and D3*D4 can be transformed into DT, why can’t D1*(DT basing D2) be rewritten as (DT basing D2)*D1, and DT basing D2 transformed as something else?
(DT basing D2)*D1 basically means ((10*10)/2)c1*10c6*10c4, and mathematically it doesn’t make a difference, but logically it translates as you choose 2 black balls first and decide to put them in 1 or 2 urns, filling accordingly, and then choose your arrangement with the rest of the 10 balls.
I hope this expands your view that choosing 2b = choosing 6w4b, choosing 2w = choosing 6b4w, choosing 1b1w = choosing 5w5b. If you can imagine (10*10)/2 you can definitely imagine this. So why do you always approach with 6w4b, 5w5b and 4w6b in the front?
And if they can go to the back of the formula, think about what you can do to transform DT basing on D2.
So if I am to interpret your formula as decisions to form arrangement in 6w4b, here are the decisions: Decision 1: pick 6w and 4b balls to put in 10 urns in some order. Decision 2: must put the 2 black balls in one or two urns (forced to be so as 6w6b-6w4b=2b). Decision 3: filling the first black ball and Decision 4: filling the 2nd black ball. This is gives a formula without exclude anything: 10c6*4c4*10c1*10c1. Decision 1 is reflected by 10c6*4*4c4, Decision 2 is reflected by c1, and Decision 3 and 4 basing on Decision 2 are 10c1 and 10c1.
Since Decision 3 and 4 can be made in reversed order as 10c1*10c1 = 10c1*10c1, and reduce two decision to a single decisions: ((10*10)/2)c1, and now it is exactly as yours: 10c6*4c4*((10*10)/2)c1. To avoid confusion we call this reduced decision Decision Three or Decision T = (10*10)/2. So your formula basically can be written as D1*(DT basing D2). And with the logic D3*D4 = D4*D3, and D3*D4 can be transformed into DT, why can’t D1*(DT basing D2) be rewritten as (DT basing D2)*D1, and DT basing D2 transformed as something else?
(DT basing D2)*D1 basically means ((10*10)/2)c1*10c6*10c4, and mathematically it doesn’t make a difference, but logically it translates as you choose 2 black balls first and decide to put them in 1 or 2 urns, filling accordingly, and then choose your arrangement with the rest of the 10 balls.
I hope this expands your view that choosing 2b = choosing 6w4b, choosing 2w = choosing 6b4w, choosing 1b1w = choosing 5w5b. If you can imagine (10*10)/2 you can definitely imagine this. So why do you always approach with 6w4b, 5w5b and 4w6b in the front?
And if they can go to the back of the formula, think about what you can do to transform DT basing on D2.
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You kept saying that Dr. Peterson’s approach is the same as yours, then the least you should admit is that choosing 2b first and 6w4b later or choosing 6w4b first and 2b later make exactly 0 difference to solving the problem! But his approach - the D1 after DT basing on D2 approach - D1 can be further expanded as choosing where to put 3rd ball and the rest 9 balls, or where to put the 3rd and 4th ball and the rest 8 balls and transforming DT basing on D2 accordingly. And In your approach, the visualization of filling the 10 balls in prior to the 2, it limited your options. I suggested that if you must visualize filling 10 prior to the 2, picturing something like emptying 1 or 2 urn(s) after you filled 10 urns with 1 ball each, and make another decision with 3 or 4 balls, but you kept saying “I am not emptying any urns”. If you fixate on the process of filling 1 urn with 1 ball first and 11th and 12th at the end, and how many ways are there in filling them, of course you will have more ways to go through the process and a lot of ways will result to the same arrangements as double counts, and you must come up with complicated exclusion methods.