how to generate permutations of N numbers in K positions

I want to generate all permutations of N numbers in K places, where K is less than N (nPk) in matlab, I've searched online and already existing questions but couldn't find a functions which generates such permutations without repitition. There are some functions which do this with repitation.

For example if I've a vector [1 2 3 4] my N is 4 and my K is 2, then I want 12 permutations using the formula N!/(N-K)1 = 4!/(4-2)! = 12

[1,2]

[1,3]

[1,4]

[2,1]

[2,3]

[2,4]

[3,1]

[3,2]

[3,3]

[4,1]

[4,2]

[4,3]

These are the permutations which I want to generate. Kindly suggest me the solution.

Answer 1

[Stephen23]

Download Loginatorist's powerful FEX submission combinator:

https://www.mathworks.com/matlabcentral/fileexchange/24325-combinator-combinations-and-permutations

and use it like this:

>> sortrows(combinator(4,2,'p'))
ans =
     1     2
     1     3
     1     4
     2     1
     2     3
     2     4
     3     1
     3     2
     3     4
     4     1
     4     2
     4     3
     

If you want to apply this to a vector that is not 1:N, simply use the output of combinator as indices into your vector.

Answer 2

[Bruno Luong]

No loop, no extrenal file needed

N=5; K=3;

P=nchoosek(1:N,K);
P=reshape(P(:,perms(1:K)),[],K)

P = 60×3

3 2 1 4 2 1 5 2 1 4 3 1 5 3 1 5 4 1 4 3 2 5 3 2 5 4 2 5 4 3

You might further sort the permutation so that the order is easily to follow

P = sortrows(P)

P = 60×3

1 2 3 1 2 4 1 2 5 1 3 2 1 3 4 1 3 5 1 4 2 1 4 3 1 4 5 1 5 2

[Jonathan Staebell]

Thanks! Just what I needed for a chemistry project (trying to calculate the changes in energy of all the different energy levels of hydrogen).

Jonathan

[Subzero]

Can you explain the P(:,perms(1:K)) part of your code please? I understand the : refers to all of the rows of P, but how can the second argument, perms(1:K), be a matrix? Thanks.

[Walter Roberson]

@Subzero

If you have an array P that is 2 or more dimensions (not a vector!), and you use P(A,B) where A and B might be arrays, then the result is the same as if you had used P(A(:), B(:)) -- so P(A(1),B(1)), P(A(2),B(1)), P(A(3),B(1)) up to P(A(end),B(1)) is the first column, then the second column would be P(A(1),B(2)), P(A(2),B(2)), P(A(3),B(2)) to P(A(end),B(2)), and so on -- all combinations of the elements in A with all of the elements in B, same as if A and B had been vectors rather than 2D arrays.

The shape of the result might be different if P is a vector instead of a 2D array.

[Subzero]

@Walter Roberson

Thanks for explaining that. I get it now.

Answer 3

[Sergey Kasyanov]

Hello!

Use nchoosek function with combination of perms function. That solution is slow but does not require any side files.

res = nchoosek(1:4, 2);
for i = 1:size(res,1)
    res = [res; perms(res(i,:))];
end
res = unique(res, 'rows');

[Abdelrahman Mohamed]

Thanks

Sergey Kasyanov

that is works