The question is a bit complicated and hard to follow.. Anyways I tried writing some code to the best of my understanding of the problem. Here is the part that finds the possible combinations:

```
% input matrix
A=[1 2 0 1 2 0 0 0
2 1 1 1 0 2 2 0
3 0 0 0 0 1 1 1
4 0 2 0 1 1 1 2
5 0 0 0 0 0 1 0
6 1 0 1 1 2 0 2
7 1 1 2 2 2 1 1
8 0 1 1 2 2 0 0
9 0 1 1 2 2 0 0
10 2 2 2 2 0 0 1];
% start by considering all rows
rowsIndices = (1:size(A,1))';
rIdx = true(size(rowsIndices));
% exclude 7th row (i.e rows with no zeros in columns 2:7)
%idx(~any(A(:,2:7)==0,2)) = false;
rIdx(7) = false;
% exclude rows that dont have zero in column 8
rIdx(A(:,8) ~= 0) = false;
% for each possible n-combinations
N = sum(rIdx);
combs = cell(1,N);
for k=2:N
% all combinations of k-rows
combsK = nchoosek(rowsIndices(rIdx), k);
% must involve first row
combsK = combsK(any(combsK==1,2),:);
% exclude from current k-combinations if there are smaller ones
if k > 2
combsKIdx = true(size(combsK,1),1);
for kk=2:k-1
if isempty(combs{kk}), continue, end
for i=1:size(combs{kk},1)
combsKIdx(sum(ismember(combsK,combs{kk}(i,:)),2)==kk) = false;
end
end
combsK = combsK(combsKIdx,:);
end
% for every possible combination, each column 2:7 must not be all zeros
combsKIdx = true(size(combsK,1),1);
for i=1:size(combsK,1)
combsKIdx(i) = all(any(A(combsK(i,:),2:7),1));
end
combsK = combsK(combsKIdx,:);
% store combinations found
combs{k} = combsK;
end
% display results
celldisp(combs)
```

Here are the combinations I got:

```
combs{1} =
[]
combs{2} =
1 2
combs{3} =
1 5 8
1 5 9
combs{4} =
[]
combs{5} =
[]
```

in other words three combinations; first with rows [1 2], second [1 5 8], and third with rows [1 5 9].

The part I left out is the final step of computing the “scores” of each combination found. Honestly I didn’t understand how, the description was confusing! So I’ll leave that part to you..

7

solved Finding subsets of a set of vectors that fulfill some conditions