q43

You have 12 identical urns and 6 white and 6 black balls to choose from. Now you choose two urn and fill 4 balls in these two and requirement is each urn contains at most 3 balls. After this label 1 to 10 on the rest of 10 urns. Then Choose 2 urns after labeling, remember their numbers, and destroy these 2 urns. Now labeling the number you remembered on the 2 urns you chose previously. Then put the rest 8 balls in the rest of 8 urns 1 in each urn. How many ways to do this so the 10 urns that are not destroyed can form an arrangement which is unique from one another? This gives you the same result as our original problem. If you approach with the exact sequence of events I described, it would be a bit more complicated than Dr. Peterson’s but a lot simpler than yours.
 
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sky2rain said:
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.
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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

Yes you understood my approach . Now what are you expecting from me?
sky2rain said:
So why do you always approach with 6w4b, 5w5b and 4w6b in the front?
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Thats how i approached . What's wrong?
 
Saumyojit said:
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

Yes you understood my approach . Now what are you expecting from me?

Thats how i approached . What's wrong?
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By putting the term 10c6*4c4 in the front of formula is where the wrong begins. Maybe I wasn’t clear enough. If you understand that 11th fill is the same as 12th fill for 6w4b, that dividing by 2 doesn’t just mean you can reverse the decision on the balls, but on the urns. In the sequence of events you chose a black ball on 11th fill and a black ball on 12th fill, and by reversing that decision it doesn’t only mean you can pick the ball on the 12th fill for your 11th, since they’re not distinct. More importantly if you decide to fill in the number 6 urn on the 11th or on the 12th for the number 9, it’s exactly the same as if you filled number 9 urn on the 11th and number 6 on the 12th fill, since there you see two sets (6,9) and (9,6) to be exactly the same so you divide them by 2.

Hence you don’t need to make your 1st - 10th fill in the beginning. Not just because balls aren’t distinct, but you would limit the pattern forming in your thinking process with limited choices for the 2 balls left.
 
The pattern forming in this problem is dependent to 4 balls you put in 2 urns, not the first 10 balls in 10 urns and filling the last 2 balls at last for 1 or 2 urns. There actually isn’t any 11th fill or 12th fill in this problem. But to show you what you did wrong I needed to say 11th and 12th and 1st - 10th fill by following your thinking.

It’s 10 urns and 10 fills, and that’s all. There’s no sequence in how you fill the urns so there’s no 1st, 4th or 10th. What happens first and last don’t make a difference… so the wrong is you kept separate them by visualizing you have to fill 10 first and 11th and 12th later. That’s why you insist on putting the 10c6*4c4 in front.
 
If you put that term to the back you would begin to see there is no 10c6*4c4, for that means 6 identical items in 6 out of 10 distinct urns and the rest are identical items put in other urns, as a final consequence, not something transiently happened. There is nothing transiently happened in a discrete problem and when you have 12 equal amount of 2 kinds of identical items to fit storage space of 10 with the requirement that there’s no 0 item storage, there is no such case that 6 identical items can be stored in 6 storages and the rest holds everything as identical items. That’s essentially what’s wrong.
 
Even if you stand by your 12 fills logic, and cognitively understand that decision sequence can shift freely, if you fill the 1st, 2nd and 3rd to be all black balls in 1 urn and the rest as 4th to 12th (notice that is 9 fills, not 8 or 10), you would write something like 10c1*1c1*1c1*9c6*3c3, because out of your 12 fills with one urn containing 3, you need to fill the SAME urn (that why 1c1) in order to achieve that and when finished you cannot use that urn anymore (that’s why 9c), so 10c6 wouldn’t exist even if you still have 6 white balls, the urns has only 9 left.

And mathematically you swap that term to the front it wouldn’t matter, but logically it should mean that you chose 6 urns out of 9 to fill white and 3 out of the rest to fill black and leave one RESERVED for your FINAL fill, not still trying to decide what can go where. Because your Unique fill, regardless if you shift the decision to the 1st 2nd and 3rd fill or 4th 7th and 9th fill, they all GO TOGETHER in ONE URN and take up that urn, and that urn is NOT to be filled with anything else and its content DISTINCT from the content of whatever other urn holds.

So if swapping it in front it should be (10c9*(9c6*3c3)*1c1)*1c1*1c1. This would best fit the sequence if you insist on filling 1st to 10th first and 11th and 12th later.
 
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As I said earlier the solution is dependent to filling 4 balls in 2 distinct urns with at least 1 in a urn. The more standardized formula should be (if 3 of the same color in 1 urn and 1 of the same color as the 3 balls in 1 in another urn) : 10c2*((2c1*1c1*1c1)*1c1)*8c8*8c6*2c2. The 9c formula is a short cut. Less number will turn up in each individual 8c formula but this would form more patterns since you are considering how to put 4 balls in 2 urns to begin with, but the final will be the same.
 
6white , 4 black balls--> 10c6 * 4c4 * (10*10)/2 this simply means you put 6 white balls and 4 black balls in 10 urns numbered 1-10, and write down 2 urn numbers on a piece of paper, and how many arrangements there exist for 10 urns and this piece of paper, provided the number can repeat but the sequence of the 2 numbers don’t matter. Doing this didn’t alter the content in any of the 10 urns. If you alter the content in any of the 10 urns, the original arrangement in 10c6*4c4 wouldn’t stand at all.

P.S. even if the question asks you to pick two numbers of urns, you shouldn’t do (10*10)/2, but ((10*10-10)/2)+10 = 55, not 50, since you didn’t double count the 00, 11, 22, 33… 99, so you need to minus these 10 pairs, then divide by 2, then add the 10 back.
 
sky2rain said:
6white , 4 black balls--> 10c6 * 4c4 * (10*10)/2 this simply means you put 6 white balls and 4 black balls in 10 urns numbered 1-10, and write down 2 urn numbers on a piece of paper, and how many arrangements there exist for 10 urns and this piece of paper, provided the number can repeat but the sequence of the 2 numbers don’t matter. Doing this didn’t alter the content in any of the 10 urns. If you alter the content in any of the 10 urns, the original arrangement in 10c6*4c4 wouldn’t stand at all.

P.S. even if the question asks you to pick two numbers of urns, you shouldn’t do (10*10)/2, but ((10*10-10)/2)+10 = 55, not 50, since you didn’t double count the 00, 11, 22, 33… 99, so you need to minus these 10 pairs, then divide by 2, then add the 10 back.
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10 distinct urns are there, say 1,2,....10. Each urn must have at least one ball.
Case 1 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are distributed in any two urns.
6 urns can be selected in 10C6 ways.
Put one white ball in each of the selected 6 urns. Since the white balls are identical, this can be done only in 1 way
4 urns can be selected from the remaining 4 urns in 4C4 ways .
Put one black ball in each of the selected 4 urns. 1 way of doing this
Select any two urns in 10C2 ways
Place two black balls in these urns. 1 way of doing so
Total number of ways = 10C6 × 4C4 × 10C2
Case 2 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are distributed in any two urns
Similar to what we have seen in case 1, this can also be done in 10C6 × 4C4 × 10C2 ways
But case 2 and case 1 together has resulted in overcounting of some of the possibilities . Suppose, in case 1, a white ball is placed in each of the 1-6 urns, a black ball is placed in each of the 7-10 urns and remaining two black ball is placed in 5 and 6, as given below
1 2 3 4 5 6 7 8 9 10
W W W W W W B B B B
B B
Assume, in case 2, a black ball is placed in each of the 5-10 urns, a white ball is placed in each of the 1-4 urns and remaining two white balls are placed in 5 and 6 urns,as given below
1 2 3 4 5 6 7 8 9 10
W W W W B B B B B B
W W
As we observe, such possibilities which are already counted in case1, have been counted in case 2 as well. This overcounting occurs in case 2, when the last two white balls are placed in urns which has a black ball. Count of such possibilities are 10C6 × 4C4 × 6C2. This need to be reduced from the total to balance the effect of overcounting.
Case 3 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are placed together in any one urn.
6 urns can be selected in 10C6 ways.
Put one white ball in each of the selected 6 urns. 1 way of doing this
4 urns can be selected from the remaining 4 urns in 4C4 ways .
Put one black ball in each of the selected 4 urns. 1 way of doing this
Select any one urn in 10C1 ways
Place two black balls in these urns. 1 way of doing so
Total number of ways = 10C6 × 4C4 × 10C1
Case 4 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are placed together in any one urn
As seen in case 3, this gives another 10C6 × 4C4 × 10C1 possibilities
Unlike what happened in case 1 and case 2, no overcounting occurs here
Case 5 : Any of the 5 urns contain one white ball each. Rest of the 5 urns contain one black ball each. Remaining one white ball is placed in any of the 5 urns which already has a white ball. Remaining one black ball is placed in any of the 5 urns which already has a black ball.
(Note that if we place the last white ball and/or the last black ball in urns where a different colour ball is already placed, it will result in overcounting of what we have already taken in case 1, case 2, case 3 and case 4)
5 urns can be selected in 10C5 ways.
Put one white ball in each of the selected 5 urns. 1 way of doing this
Remaining 5 urns can be selected in 5C5 ways.
Put one black ball in each of these 5 urns. 1 way of doing this
Select any one urn which has a white ball in 5C1 ways
place the remaining white ball in this urn. 1 way of doing this
Select any one urn which has a black ball in 5C1 ways
place the remaining black ball in this urn. 1 way of doing this
Total number of ways = 10C5 × 5C5 × 5C1 × 5C1
Therefore, required number of ways
= (10C6×4C4×10C2)+(10C6×4C4×10C2)-(10C6×4C4×6C2)+(10C6×4C4×10C1)
+(10C6×4C4×10C1)+(10C5×5C5×5C1×5C1)
= (10C6×10C2)+(10C6×10C2)-(10C6×6C2)+(10C6×10C1)+(10C6×10C1)+(10C5×5C1×5C1)
= 2(10C6×10C2)-(10C6×6C2)+2(10C6×10C1)+(10C5×5C1×5C1)
= 2(210×45)-(210×15)+2(210×10)+(252×5×5)
= 18900 - 3150 + 4200 + 6300
= 26250
 
what happens if you place the balls in front of the urns instead of inside the urns? Just line them up and put them in 1 time instead of 2 times or 12 times? Does this change any consequence? How would I understand your formula then?

I think this question is more about line up 12 balls as 9+3 or 8+2+2, and rearrange urns to store them, not really rearranging balls…

you can line them up in one way: wwwww bbbb wbb, and rearranging 10 urns for their storage (but not actually storing them), so by different urns to put to store the same balls, as long as one urn’s ball-in-front is different from previous arrangement, it’s a new arrangement.
 
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Or better yet let me ask this question: 10 identical urns, their known content is 12 balls in them, 6 white and 6 black and at least 1 ball in each urn. Question 1: how many possible patterns do the 12 balls form in 10 identical urns and Question:2 If you want to label the ten urns number 1 to number 10, and the urns become distinct, what total ways are there to label for all possible patterns?


How would I understand your formula then? So your formula is only logical for one problem, not all problems that can give the same final result.
 
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sky2rain said:
Or better yet let me ask this question: 10 identical urns, their known content is 12 balls in them, 6 white and 6 black and at least 1 ball in each urn. Question 1: how many possible patterns do the 12 balls form in 10 identical urns and Question:2 If you want to label the ten urns number 1 to number 10, and the urns become distinct, what total ways are there to label for all possible patterns?
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Let me just do one pattern: wwwwww +bbb+bbbin1. If you do 10c6, the rest will be 4c3*1c1, so 10c6*4c3*1c1=840. And if you do bbbin1 first, 10c1*9c6*3c3, it still = 840. And if you do bbb first, 10c3*7c6*1c1, it still = 840. So 10c6*4c4*10c2 doesn’t make any sense at all… basically you created so many duplicates yourself by understanding the problem wrongly and then worked so hard to figure out how to exclude them. The number 10 should only appear once in your formula!
 
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