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

Hi all,

I have a series of number and a single number. Is there any way to find out how many exist to build single number by summation of series members?

for example, suppose the series is: 1, 2, 4, 5

and the single number is 6 there are two ways (2+4) and (5+1)

it is easy when the numbers in series are identical.

for example if series is: 1, 1, 1, 1 and the number is 2 the number of possible ways is 4!/(2! * (4-2)!)

but I like to have a general way by which the possible ways can be calculated even by existing a series with different numbers.

Thank you in advance for your help:)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,433

Hi;

Yes, it is possible to calculate those partitions but first I must ask how do you get 6 from the set of {1,1,1,1}? Looks like only 1+1 = 2.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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

I'll be grateful if you provide some information about the method.

it is impossible to make 6 form {1, 1, 1, 1}; in second example the desired number was 2.

for example if series is: 1, 1, 1, 1 and <the number is 2> the number of possible ways is 4!/(2! * (4-2)!)

the 6 was for first example.

excuse me for my bad English.

thank you for your reply

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,433

Do you mean greater than 2? You would have to use more than 2 numbers to do that.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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

lets explain it from beginning. it has been confusing.

I have a set of numbers named A, and it has at least 4 numbers. I also have a number (lets call it X).

X is always smaller or equal to summation of A members.

I like to know how many ways exist to build X from A members.

when all member of A are identical it is easy for example A={1, 1, 1, 1} and X=3 the number of possible ways can be calculated by 4!/(3! 1!) = 4 ways.

is there any way to formulate this for sets that have not identical members for example A={ 1, 2, 3, 7, 9...}?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,433

Hi kappa_am;

For the set {1,1,1,1} there is only one way to get 3, 1+1+1 = 3 and one way to get 4, 1+1+1+1 = 4. That is two ways. How do you get 4 ways?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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

it is important which 1 takes part in building 3. {1st , 2nd, 3rd}, {2nd, 3rd, 4rt}, {1st, 2nd, 4th}, {1st, 3rd, 4th}

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,433

You are treating the ones as being different?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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

yes, how ever their value are identical but their position in the set is important.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,433

Hi;

And you want greater than or equal to x?

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

exactly equal.

just I know X is never greater than summation of all members of the set.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,433

I will see if I can get something. Post here if I do.

(2+4) and (5+1)

Does order of the others count?

Are (4+2) and (1+5) solutions too?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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

the order of number in the choice doesn't count. for example in first example where the set was A= {1, 2 ,4 ,5} and the X=6 there are 2 possible ways (2+4) and (5+1). the transposed answer doesn't count as new way. it means (4+2) and (2+4) counts just 1 way.

but if there was another 4 in series A={1,2,4,4,5} there were 3 possible ways.(1+5), (2+4(1st)), (2+4(2nd)

thank you for your help

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,433

So far I can with the help of generating functions do some of the problem.

It looks like it is possible to count them but a formula for it is going to depend on what you want.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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

I dealt with it too much and really exhausted. could you introduce some useful resources that provide me some information to solve this problem?

Thank you

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,433

A general formula for it is going to have to take into account the number and the types of elements that are in the set. Since they are going to vary a lot I would say a formula as I understand it is probably out.

I can solve for a specific case using generating functions and if you are up to it you can start there. You may not solve the problem but you will pick up a lot of useful mathematics.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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

thank you for your help I hadn't known about generating functions. I have find some text on net. it will be my fortune if you answer my question probable may rise after reading them.

thank you

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 106,433

I keep looking for a solution to it and who knows...

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**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 135

In the example where the elements are the same, such as {1,1,1,1}, and you have X=3, the number of ways to get X=3 is simply 4C3. Very interesting problem.

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