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**samuel.bradley.99****Member**- Registered: 2016-03-16
- Posts: 51

Given a set {1,2,…,14}, find all possible arrangements of the set for which floor(n/2) precedes n, for any n∈(1,14].

FYI floor(n/2) = integer part (n/2)

*Last edited by samuel.bradley.99 (2016-03-16 01:53:12)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Can I see an example?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**samuel.bradley.99****Member**- Registered: 2016-03-16
- Posts: 51

1,3,2,5,4,8,6,7,10,9,11,12,14,13 or

1,2,5,4,3,9,10,6,8,11,12,13,7,14.

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**samuel.bradley.99****Member**- Registered: 2016-03-16
- Posts: 51

If this is of any help, I got the answer (played with the "Combinations and Permutations Calculator" and with the "pattern" tools) for n = 3 to n = 11 but the tool cannot process such big numbers for n = 14

So for n=4 the arrangements that satisfy the requirement are 3 out of the total of 24

n=5 --> 8 of 120

n=6 --> 20 of 720

n=7 --> 80 of 5040

n=8 --> 210 of 40320

n=9 --> 896 of 362880

n=10 --> 3360 of 3628800

n=11 --> 19200 of 39916800

Of course all the possible permutations for n=14 are 14! and also number 1 can only be in 1st position. I will leave the rest for you, as I am a novice at combinatorics

Can you please show the method to calculate this and the logic behind it?

I started by setting all the possible positions for each number:

1 only goes to 1st position.

2 can be in any position of 2, 3, 4 or 5.

3 in any position of 2, 3, 4, 5, 6, 7, 8 and so on.

Actually, not in "any" position - these are only the starting positions, to which then we must also apply the restriction of [n/2] preceding n in the arrangement.

But then I don't know how to continue

The rest is yours

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

The method and logic can be really simple if you can immediately answer a few more questions:

What do you get for n = 1,2,3,4?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**samuel.bradley.99****Member**- Registered: 2016-03-16
- Posts: 51

For n=4 I get 3 of 24

For n=3 I get 2 of 6

for n=2 I get 1 of 2

n=1 is out of scope, since floor(n/2)=0 and 0 does not belong to the set.

bobbym wrote:

The method and logic can be really simple if you can immediately answer a few more questions:

What do you get for n = 1,2,3,4?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hmmm, then allow me to make a nice ansatz...

n = 12 = 79200

n = 13 = 506880

n = 14 = 2745600

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**samuel.bradley.99****Member**- Registered: 2016-03-16
- Posts: 51

Haha I had to google the word ansatz

Can you share your calculations?

bobbym wrote:

Hmmm, then allow me to make a nice ansatz...

n = 12 = 79200

n = 13 = 506880

n = 14 = 2745600

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Do you agree with the answer given?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**samuel.bradley.99****Member**- Registered: 2016-03-16
- Posts: 51

Unfortunately I don't know the answer and I have only performed the calculations up to n=11 - like I said, I did them using the https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html but it can only handle numbers up to 10000000

Does your method give the same results with mine, for n=2 to n=11? By the way, these are kind of "confirmed" because I also did the calculations in Excel, but I would like to see a more "scientific" way, with arrangements, permutations etc.

bobbym wrote:

Do you agree with the answer given?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Yes, I am getting the same answers for other n.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 232

Wow!! Heap sort!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Heaps, is more accurate I guess.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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