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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

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: 87,237

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

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: 87,237

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

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: 87,237

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.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

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: 87,237

You are treating the ones as being different?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

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: 87,237

Hi;

And you want greater than or equal to x?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

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: 87,237

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?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

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: 87,237

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

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: 87,237

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**kappa_am****Member**- Registered: 2013-12-20
- Posts: 35

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: 87,237

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 120

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.

The eclipses from Algol (an eclipsing binary star) come further apart in time when the Earth is moving away from Algol and closer together in time when the Earth is moving towards Algol, thereby proving that the speed of light is variable and that Einstein's Special Theory of Relativity is wrong.

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