how many possibility exist? (discrete mathematics)

kappa_am · 2014-03-19 07:55:27

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:)

bobbym · 2014-03-19 08:07:09

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.

kappa_am · 2014-03-19 23:40:45

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

bobbym · 2014-03-20 01:39:13

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

kappa_am · 2014-03-20 02:33:20

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...}?

bobbym · 2014-03-20 02:39:44

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?

kappa_am · 2014-03-20 02:46:04

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

bobbym · 2014-03-20 02:47:14

You are treating the ones as being different?

kappa_am · 2014-03-20 02:56:52

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

bobbym · 2014-03-20 03:00:16

Hi;

And you want greater than or equal to x?

kappa_am · 2014-03-20 03:02:37

exactly equal.

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

bobbym · 2014-03-20 03:09:27

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?

kappa_am · 2014-03-21 00:48:05

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

bobbym · 2014-03-21 01:52:39

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.

kappa_am · 2014-04-06 05:55:24

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

bobbym · 2014-04-06 06:00:30

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.

kappa_am · 2014-04-06 06:24:05

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

bobbym · 2014-04-06 06:25:00

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

Fruityloop · 2014-04-08 15:04:42

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.

Math Is Fun Forum

#1 2014-03-19 07:55:27

how many possibility exist? (discrete mathematics)

#2 2014-03-19 08:07:09

Re: how many possibility exist? (discrete mathematics)

#3 2014-03-19 23:40:45

Re: how many possibility exist? (discrete mathematics)

#4 2014-03-20 01:39:13

Re: how many possibility exist? (discrete mathematics)

#5 2014-03-20 02:33:20

Re: how many possibility exist? (discrete mathematics)

#6 2014-03-20 02:39:44

Re: how many possibility exist? (discrete mathematics)

#7 2014-03-20 02:46:04

Re: how many possibility exist? (discrete mathematics)

#8 2014-03-20 02:47:14

Re: how many possibility exist? (discrete mathematics)

#9 2014-03-20 02:56:52

Re: how many possibility exist? (discrete mathematics)

#10 2014-03-20 03:00:16

Re: how many possibility exist? (discrete mathematics)

#11 2014-03-20 03:02:37

Re: how many possibility exist? (discrete mathematics)

#12 2014-03-20 03:09:27

Re: how many possibility exist? (discrete mathematics)

#13 2014-03-21 00:48:05

Re: how many possibility exist? (discrete mathematics)

#14 2014-03-21 01:52:39

Re: how many possibility exist? (discrete mathematics)

#15 2014-04-06 05:55:24

Re: how many possibility exist? (discrete mathematics)

#16 2014-04-06 06:00:30

Re: how many possibility exist? (discrete mathematics)

#17 2014-04-06 06:24:05

Re: how many possibility exist? (discrete mathematics)

#18 2014-04-06 06:25:00

Re: how many possibility exist? (discrete mathematics)

#19 2014-04-08 15:04:42

Re: how many possibility exist? (discrete mathematics)

Board footer