combinations of a matrix.....

I am trying to find an command/algorithm which gives me all possible translates of a given matrix that has fixed #a,#b, and #c. By translates I mean all those matrices which have the same #a,#b, and #c but different locations for these elements .

For example

M=[3 3 3;1 2 3;1 2 3]

the answer would be something like

M1=[1 2 3;1 2 3;3 3 3]
M2=[1 1 3;2 2 3;3 3 3]....

I dont know if its possible for matrices or not. Should I think of matrix as an array and try to find combinations for that array.

If there exists such an algorithm then kindly name it so that I can look into its theory

[Walter Roberson]

All of the unique permutations of the same sets of elements?

[xplore29]

Its like you first find the numbers of times 1,2, and 3 in the given example and then find all those matrices which have the same number of 1s, 2s, and 3 . So the answer will be set of all matrices which are different.I didn't know whether i should be calling them permutations of a given matrix or what.

[xplore29]

So yes you may call it unique permutations of the same set of elements

Answer 1

[Roger Stafford]

If I understand you correctly, your problem is the same whether your M is a matrix or a single vector with the same total number of elements, so for convenience I will assume you really wish to fill a row vector with your three numbers in all possible combinations. If there are to be na, nb, and nc counts of the three quantities a, b, and c respectively, taken in all possible combinations, then the total number of these combinations would be (na+nb+nc)!/na!/nb!/nc!. Our task is therefore to generate a single matrix M with that many rows and na+nb+nc columns where each row is one of the possible combinations. (You should be cautious about how large you make these counts. The total number of combinations can be a surprisingly large number.)

We can do this with two calls on matlab's 'nchoosek' function, and we will take advantage of the fact that

 (na+nb+nc)!/na!/nb!/nc! = (na+nb+nc)!/na!/(nb+nc)! * (nb+nc)!/nb!/nc!

which as you see is the product of two particular 'nchoosek' combination count numbers.

The following matlab code should accomplish this goal. (There is probably a more efficient way of generating this but I am too lazy at the moment to work it out.)

 C1 = nchoosek(1:na+nb+nc,na); % Here are the two calls on 'nchoosek'
 C2 = nchoosek(1:nb+nc,nb);
 M = c*ones(size(C1,1)*size(C2,1),na+nb+nc); % First, fill M with all c's
 k1 = 0;
 for k2 = 1:size(C1,1)
   p1 = C1(k2,:); % Use C1 to generate indices for inserting a's
   q1 = 1:na+nb+nc;
   q1(p1) = [];  % These indices will point to b and c locations
   for k3 = 1:size(C2,1)
     p2 = C2(k3,:); % Use C2 to generate indices for inserting b's
     k1 = k1 + 1; % Advance the row index of M
     M(k1,p1) = a; % Insert na a's
     M(k1,q1(p2)) = b; % and nb b's
   end
 end