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If we have 9 black blocks that stick together from 1-9 respectively to form a line and have another 9 blue blocks from 1-9 to put on a new line stick with the first line from 1-9. Find the probability of how you could stick the blue block with the black blocks so that the number won't be the same.

I dunno if what I'm doing is right or wrong at all but here is my idea. All the possible move could be $(9!)$ and the possible that we could do is 8[(8×6)+(7×5)+(6×4)+(5×3)+(3×1)]. The reason why I say that is because I think that the numbers can not be swap with neighbors ( sorry I'm not good at explaining ). Correct me if I'm wrong ( I dun think I could be right at all hahahah ) I hope everyone here could help me. Thank you

Here is the picture explanation

user635988
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    Can you clarify what it means "so that the number won't be the same."? –  Dec 02 '19 at 15:03
  • You seem to be mixing up probability and number of ways – Arnaud Mortier Dec 02 '19 at 15:04
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    I don't understand what you mean by "stick the blue block with the black blocks." Are we making a tower of blue blocks, putting this to the side, then making a tower of black blocks, lining these up next to one another and comparing the numbers for each block at each level of the tower and asking if there are any matches? Or equivalently, taking one blue and one black block, combining them and setting aside, and repeating this, asking if any of these many two-block combinations have matching numbers? – JMoravitz Dec 02 '19 at 15:04
  • What numbers are you forming? Are they $18$ digits, formed from distinct sets of numbers $1$ to $9$? It doesn't look like you are calculating a probability, you are trying to calculate how many numbers you can form. There don't seem to be any moves. Please clarify the question. – Ross Millikan Dec 02 '19 at 15:05
  • Also could you clarify the "nice" from the title? So far, it's only unclear. – Arnaud Mortier Dec 02 '19 at 15:05
  • @M.Nestor it is something like the first line is 1-2-3..-9 and then you have another nice numbers to form a new line that will stick together with the first one but you couldn't put the same number on the same line...I mean like Sudoku – user635988 Dec 02 '19 at 15:06
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    So, exactly my first interpretation then? The answer will be $\dfrac{!9}{9!}$ then. Take a look at derangements and note that without loss of generality the blue tower might as well have been in the order $1,2,3,\dots,9$ and we only randomly arrange the black tower. – JMoravitz Dec 02 '19 at 15:07
  • So basically you are asking for the total number of possible patterns the first two rows of a sudoku grid can have? – Arnaud Mortier Dec 02 '19 at 15:07
  • @ArnaudMortier yeah it could be something like that but we already know about the first line ( 1-9 respectively) so the second line is blank. – user635988 Dec 02 '19 at 15:11
  • Then you've got the answer as given by @JMoravitz. – Arnaud Mortier Dec 02 '19 at 15:13
  • I put the link to the photo ( I can't upload photo directly sorry ) – user635988 Dec 02 '19 at 15:19
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    Possible duplicate of I have a problem understanding the proof of Rencontres numbers (Derangements) – JMoravitz Dec 02 '19 at 15:21
  • If you are counting the number of ways to fill out the rest of the spaces, then the answer is $!9$, also written $\mathcal{D}_9,\mathcal{D}(9)$ and a few other ways. I included a link to the wikipedia article above already. The question I linked to contains the definition as well as a number of recurrences and proofs of formulas. The most accessible proof to a beginning student is usually the one involving inclusion-exclusion and is available in both links I provided. If asking for the probability, then divide the result by $9!$. Don't confusing counting with probability. – JMoravitz Dec 02 '19 at 15:24
  • @JMoravitz yeah i know it's 9! For the possible way to put it. But what I want is to put the number on block so that it is not the same ( if you dun mind could you take a look at my picture I just upload at the moment it could be helpful to understand what I want ) – user635988 Dec 02 '19 at 15:28
  • Do not confuse the subfactorial $!9=133496$ which counts the number of derangements and the factorial $9!=362880$ which counts the number of permutations. These are different numbers. Slowly and carefully read everything else I've already said. – JMoravitz Dec 02 '19 at 15:30
  • @JMoravitz oops sorry about that but did you understand my question yet?:( If u dun mind could you simplify my question, I'm not good at explaining it to other – user635988 Dec 02 '19 at 15:32
  • Yes, I understood the question from the very beginning and had immediately told you that there are $!9$ ways of arranging the blocks so that no numbers match and the probability of having done so is $\dfrac{!9}{9!}$. Again, go back and reread everything I've already said so far, there is nothing more to talk about and I would just be repeating myself at this point. – JMoravitz Dec 02 '19 at 15:33
  • @JMoravitz thank you so much I'll go research about Derangement more – user635988 Dec 02 '19 at 15:35

1 Answers1

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A = the numbers aren't the same

Calculating the total number of outcomes:

Card(Ω) = n! = 9! = 362880

The number of favorable outcomes is the nearest integer to n!/e

Card(A) = n!/e = 9!/e = 133496

P(A) = Card(A)/Card(Ω) = 133496/362880 = 0.36787918871

Lioneds
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