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## #1 2014-04-07 06:50:26

kappa_am
Member
Registered: 2013-12-20
Posts: 55

### Permutation

I have a number named X and N number of empty rooms. in how many ways I can fill this room to have X? all rooms have to be filled and all smaller number of N should be obtained by addin some of these numbers. and I know x is greater than N. is there any formula?
Example X=9 and N=4 there are the possibilities:
1,1,3,4; 1,1,4,3; 1,3,1,4; 1,3,4,1; 1,4,1,3; 1,4,3,1; 1,2,24; 2,2,1,4; 2,2,4,1; 4,2,2,1; 1,1,2,5;1,1,5,2; 5,1,1,2; 5,1,2,1; 2,1,1,5; 2,1,5,1; 2,5,1,1; 5,2,1,1,

thank you:)

Last edited by kappa_am (2014-04-07 07:47:45)

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## #2 2014-04-07 07:28:37

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

### Re: Permutation

Hi kappa_am;

This is similar to the other question. Can we use the natural numbers up to X?

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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## #3 2014-04-07 07:36:49

kappa_am
Member
Registered: 2013-12-20
Posts: 55

### Re: Permutation

I have modified the question. I have missed some information sorry.
all natural numbers can be used. gust there are 2 rules the numbers have to add  up to X and all smaller number [1,X-1] could be obtained by adding some of numbers in series. to fulfill this I think the largest number have to be at max first integer number larger than X/2 for X=9-->largest number 5.

Last edited by kappa_am (2014-04-07 07:50:11)

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## #4 2014-04-07 07:54:32

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

### Re: Permutation

I am getting

for 9 and 4. Any members you do not think belong there? If you agree with the two lists then we have a formula!

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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## #5 2014-04-08 02:15:13

kappa_am
Member
Registered: 2013-12-20
Posts: 55

### Re: Permutation

there is a condition all numbers up to x=9 could be obtained using the series member but: we cannot generate 4 using 6,1,1,1 or I cannot generate 1 using 2,2,3,2...
some members have to be obliterated.
according to above condition at least one room has to be 1.

Thank you for taking time to help me.

Last edited by kappa_am (2014-04-08 03:10:35)

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## #6 2014-04-08 03:31:48

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

### Re: Permutation

I am not following you. How can you ever generate 1 using 4 integers all greater than 0?

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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## #7 2014-04-08 03:55:19

kappa_am
Member
Registered: 2013-12-20
Posts: 55

### Re: Permutation

sorry, I write it vague.
let me clarify:
I have X and N number of rooms
here are the conditions:
summation of all N number have to be X
and all integer number less than X ([1 X-1]) should be obtained using at least one subset of those numbers.
I can obtain 1 of any series that have at least one 1 in that.

Last edited by kappa_am (2014-04-08 03:55:46)

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## #8 2014-04-08 03:59:27

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

### Re: Permutation

So, why can you not generate 1 from a subset of 6,1,1,1?

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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## #9 2014-04-08 04:10:30

kappa_am
Member
Registered: 2013-12-20
Posts: 55

### Re: Permutation

we can generate ,but we cannot generate 4 and 5 of this set 6,1,1,1
we cannot generate 1 of subsets that haven't any 1 like 2,2,3,2

Last edited by kappa_am (2014-04-08 04:12:48)

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## #10 2014-04-08 04:24:26

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

### Re: Permutation

But there are some in my list that can certainly get 1,2,3,4,5,6,7,8.

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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## #11 2014-04-08 04:38:34

kappa_am
Member
Registered: 2013-12-20
Posts: 55

### Re: Permutation

yes, the list is ok! just the sets that haven't any 1 and sets that have 6 have to be obliterate.
(6 1 1 1), (1 6 1 1), (1 1 6 1), (1 1 1 6), (2 3 2 2), (2 2 3 2), (2 2 2 3), (3 2 2 2) have to be obliterated.
I like a formula by having X and N, the number of sets can be obtained. it it obvious that at least one of rooms has to be 1. if we can find a rule for maximum number we may be able to find a  formula.
the highest number for this example is 5. it has relationship with N and X. I am trying to develop a rule for maximum number. I think after that we could develop a rule for obtaining the number of sets.

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## #12 2014-04-08 04:45:49

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

### Re: Permutation

So you are saying there has to be a 1 in every set? And no 6?

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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## #13 2014-04-08 04:54:44

kappa_am
Member
Registered: 2013-12-20
Posts: 55

### Re: Permutation

at least one 1 has to be in series because if it not be, I haven't any subset that its value is 1. this is general.
but 6 is for this specific example (x=9, N=4), because if 6 is in set all other member have to be 1 (minimum value). but in this situation I haven't any subset that add up to 4 and 5.

if x was 12 then 6 could be in series like (1,2,3,6).

I think if we could find a relationship for determining maximum value for each problem using N and X we could find the number of sets that satisfy the mentioned conditions.
Am I right?

Last edited by kappa_am (2014-04-08 04:56:02)

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## #14 2014-04-08 05:17:06

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

### Re: Permutation

Am I right?

I do not know. Partition problems are tough especially when you have to leave certain numbers out and put others in.

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.

Offline