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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?

(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.

we cannot generate 1 of subsets that haven't any 1 like 2,2,3,2

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.

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.

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.

**kappa_am**- Replies: 13

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:)

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

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

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

exactly equal.

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

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

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...}?

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

**kappa_am**- Replies: 18

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:)

in banker's sequence all possibilities are checked but I think my method reach to answer very soon.

please let me know your idea

thanks

bobbym wrote:

Hi;

I am able to write a algorithm that maps binary numbers to a subset, very efficiently. the problem is somewhat different.

I have found a way but it's too time consuming.

lets explain it by an example. our digit is {1,2,3,4,5} and in 3-digit category I like to find 9th number.

we now the smallest member of each category is a number with 1 in highest value digit and the lower digits values increment by 1 for example first member of 2-digit category is 12 and first member of 3-digit category is 123. Knowing this lets start solving the problem.

123

add the sequence number-1 to first digit; if it becomes more than number of our set numbers add 1 to first and second digit. do result minus 6. repeat this till the result become 0. In adding procedure treat the digits based on its location for example first digit is based on 5 and when it becomes 6 it overflows, the second digit is based on 4 and so on

back to example:

first digit is 3 + 8 = 11 -6 = 5. the number becomes 123+011=134

first digit is 4 + 5= 9 -6 = 3 . the number becomes 134+011=145

first digit is 5 + 3= 8 -6 = 2 . the number becomes 145+011=234

first digit is 4 + 2= 6 -6 = 0 . the number becomes 234+011=245

and yes the 9th number is 245

3-digit: 123, 124, 125, 134, 135, 145, 234, 235, 245, 345

I have checked with another conditions and set with higher number of figures and it is ok but in set with high number of members when I look for a number at the end of the category it becomes cumbersome. is there any better solution?

Dear Bobbym, I am able to program by Fortran and C++. I will check Banker's sequence

in example of post #5, the location is 6 and the category is 6.

about last post, we don't know the number, we just know the category and the number location in that category. an we like to find the number according to this information.

our categories are always consist of all possible numbers not a subset for example 2-digit category: 2-digit = {12, 13, 14, 15, 23, 24, 25, 34, 35, 45}

furthermore the figures always start from 1 and are successive for example {1,2,3} we never have {2,3,4} or {1,3,4}

sorry for being vague. this is my best English

for example I have five figure: 1, 2, 3, 4, 5. I can make numbers with 1 or 2 or 3 or...5 digit. the laws that must be considered in building this numbers are 1- a specific figure has not be used twice or more in a number ( maximum once can be used). 2- always higher order digits have greater value than lower order value ( I can have 123 but I never have 321 or 213). using these laws I can construct following numbers which are categorized according to its number of digits:

1-digit: 1, 2, 3, 4, 5

2-digit: 12, 13, 14, 15, 23, 24, 25, 34, 35, 45

3-digit: 123, 124, 125, 134, 135, 145, 234, 235, 245, 345

4-digit: 1234, 1235, 1245, 1345, 2345

5-digit: 12345

these are all possible numbers when I have figures 1,2,...,5

Now, I know the category number of a specific number and its number in that category. for example I know the number belongs to 2-digit category and is 6th number in that category. knowing these I have to find 24. I like to find an algorithm or a formula for this procedure.

I have to declare that number with more than 1-digit can be a vector. for example i=24 can be i(1)=4 and i(2)=2

Thank you

I have read all the page. but, beats me. I know the number of figures in each category. the problem is developing a formula or a algorithm to find a figure using its category number and a number that indicates its location in the category. for example having number of category 3 and location number 4 I have to reach to 134.

I couldn't understand how to use binomial theorem to develop required algorithm. i will be grateful if you provide more information.

**kappa_am**- Replies: 14

hi all,

I have a bunch of numbers; they are constructed of digits without repetition, and in this numbers a digit is always greater lower value digits. This numbers are categorized according to the its number of digit; in each category all possible numbers are exist.

here a simple example:

assume the numbers constructed of 1, 2, 3, 4, 5 and they are catagorized as follow according to number of digits:

1-digit: 1, 2, 3, 4, 5

2-digit: 12, 13, 14, 15, 23, 24, 25, 34, 35, 45

3-digit: 123, 124, 125, 134, 135, 145, 234, 235, 245, 345

4-digit: 1234, 1235, 1245, 1345, 2345

5-digit: 12345

the question is: how I can find a number by knowing the category and the number of its place in row. for example by knowing category number 2 and sequence number 7, I have to obtain: 25

since it becomes cumbersome at higher orders, I need to formulate the procedure. a formula or a algorithm is appreciated.

Thank you:)

I have a string of terms starts with ∑(n-m) {down limit of summation is m=2 and upper limit is m=(n-1). in this string each term is constructed according to the sub-terms of the previous term for example:

fist term is ∑(n-2) = (n-2)+(n-3)+(n-4)+(n-5)+...(n-(n-1)); the second term is ∑(n-3)+∑(n-4)!+∑(n-5)+....+∑(n-(n-1))

the third term will be summation of second side of the following equations:

in second term we have ∑(n-3) which is (n-3)+(n-4)+(n-5)...+(n-(n-1)) so one sub-term in third term is ∑(n-4)+∑(n-5)+∑(n-6)+...∑(n-(n-1))

in second term we have ∑(n-4) which is (n-4)+(n-5)+(n-6)...+(n-(n-1)) so another sub-term in third term is ∑(n-5)+∑(n-6)+∑(n-7)+...∑(n-(n-1))

...

in second term we have (n-(n-2)) which is (n-(n-2))+(n-(n-1)) so the last sub-term in third term is ∑(n-(n-1))=1

the other terms in this string are constructed as it is stated according to the previous term. In above equations "∑" means summation of the term located front of it starting from indicated number to n-1. for example ∑(n-4)= (n-4)+(n-5)+(n-6)+....+(n-(n-1).

I need a single formula that enables me to obtain each term. an algorithm or a program in C language is also appreciated.

Thank you very much

Sorry,

I mean summation, not multiplication.