how to obtain the number of cases of length. but those matrices sum must be 30 or something value(on constraints)

first, -constraints-

       elements must be between 4 and 12

for example,

A=[6 7 8 9],  sumA=sum(A)=30;

What is number of cases of length that can be obtained so that the sum becomes 30?

For reference,

If the length of the matrix is 5, A is [5 5 6 7 7].

[José-Luis]

Instead of asking this question again, you could have answered to the requests for clarification in this question.

When you don't, the effort of the people trying to help you is wasted.

[Stephen23]

Duplicate:

https://www.mathworks.com/matlabcentral/answers/355684-i-want-to-obtain-number-of-cases-matrix-s-length-by-applying-same-sum-in-any-matrix-on-constrai

[Walter Roberson]

Stephen, you should not close duplicate questions that have Answers unless the Answers also appear in the original location.

Answer 1

[KSSV]

x0 = 4 ; x1 = 12 ;
x = x0:x1 ;
iwant = cell([],1) ;
count = 0 ;
for i = 1:length(x)
    A = nchoosek(x,i) ;
    thesum = sum(A,2) ;
    if any(thesum==30)
        count = count+1 ;
        iwant{count} = A(thesum==30,:) ;
    end
end

You have the possible combinations whose sum is 30 in iwant.

[Brian Kim]

If the number of cases is up to 20 or any value,how to modify to code?

And in above case, number of cases are originally 5 (length are 3,4,5,6 and 7).

for example, if length is 7, A is [4 4 4 4 4 4 6].

[KSSV]

You need to only provide x.....no changes required.....

[Brian Kim]

ah~ i am sorry. i mean, some of elements can be same.

Answer 2

[Walter Roberson]

See https://www.mathworks.com/matlabcentral/fileexchange/12009-partitions-of-an-integer

[Walter Roberson]

If you just need to calculate the number of different ways that you can partition a number with particular (possibly repeated) integers, then see the mathematics outlined at https://math.stackexchange.com/questions/15521/making-change-for-a-dollar-and-other-number-partitioning-problems