Math Is Fun Forum

samuel.bradley.99

Member

Registered: 2016-03-16

Posts: 51

Integers arrangements

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)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integers arrangements

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.

samuel.bradley.99

Member

Registered: 2016-03-16

Posts: 51

Re: Integers arrangements

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.

samuel.bradley.99

Member

Registered: 2016-03-16

Posts: 51

Re: Integers arrangements

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

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integers arrangements

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.

samuel.bradley.99

Member

Registered: 2016-03-16

Posts: 51

Re: Integers arrangements

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.

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?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integers arrangements

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.

samuel.bradley.99

Member

Registered: 2016-03-16

Posts: 51

Re: Integers arrangements

Haha I had to google the word ansatz
Can you share your calculations?

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

n = 12 = 79200
n = 13 = 506880
n = 14 = 2745600

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integers arrangements

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.

samuel.bradley.99

Member

Registered: 2016-03-16

Posts: 51

Re: Integers arrangements

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.

Do you agree with the answer given?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integers arrangements

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.

anna_gg

Member

Registered: 2012-01-10

Posts: 232

Re: Integers arrangements

Wow!! Heap sort!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integers arrangements

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.