# fill matrix with all options of successive, increasing numbers 1-5

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Sjoukje de Lange on 25 Jan 2021
Commented: Sjoukje de Lange on 26 Jan 2021
I want to construct a matrix, m x n, filled with the numbers 1 till 5. The numbers have to be successive, and have to be increasing. One column of the matrix could for example look like.
A =1 1 2 2 2 2 3 4 5 5 5 5
or
A = 1 2 3 3 3 4 5 5 5 5 5 5
I want to construct a matrix with all possible options. The first column of matrix A therefore should look like this:
A(:,1) = 1 1 1 1 1 1 1 1 2 3 4 5
and the last one like this
A(:,end)= 1 2 3 4 5 5 5 5 5 5 5
In my case, the matrix will have a length of 96, instead of the above example where the length is 12. Could you help me?
Sjoukje de Lange on 25 Jan 2021
Hmmm that might be a but extreme yeah... I need repeating values because I want to label each parameter at each kilometer with number 1-5.

Bruno Luong on 25 Jan 2021
Edited: Bruno Luong on 25 Jan 2021
p = 5;
n = 12;
j = nchoosek(2:n,p-1);
m = size(j,1); % == nchoosek(n-1,p-1) == 330 and not 96
i = repmat((1:m)',1,p-1);
A = cumsum(accumarray([i(:) j(:)],1,[m n]),2)+1
Sjoukje de Lange on 26 Jan 2021
neat! Amazing!

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