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how to obtain the number of cases of length. but those matrices sum must be 30 or something value(on constraints)
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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].
3 个评论
Walter Roberson
2017-9-13
Stephen, you should not close duplicate questions that have Answers unless the Answers also appear in the original location.
回答(2 个)
KSSV
2017-9-13
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.
3 个评论
Walter Roberson
2017-9-13
1 个评论
Walter Roberson
2017-9-13
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
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