How can I obtain all possible combinations of a given vector

I have a vector A=[1 1 1 0 0 1 1]. I want all the possible combinations as following....

A=[1 1 1 0 0 0 0; 1 1 0 0 0 1 0; 1 1 0 0 0 0 1; 0 1 1 0 0 1 0; 0 1 1 0 0 0 1; 0 1 0 0 0 1 1]

[Wupadrasta Santosh Kumar]

Use matlab function "perms"

P = perms(v)

What I see is that all combinations in

A=[1 1 1 0 0 0 0; 1 1 0 0 0 1 0; 1 1 0 0 0 0 1; 0 1 1 0 0 1 0; 0 1 1 0 0 0 1; 0 1 0 0 0 1 1]

has 3 ones and 4 zeros. So, I guess initial vector A = [1 1 1 0 0 1 1] would be a mistake. If I understood right, your asnwer would be

A = [1 1 1 0 0 1 1];
As = perms(A);
RequiredA = unique(As,'row's);

[subhashree priyadarshini]

Thank you sir. But my initial vector is A=[1 1 1 0 0 1 1]. and its binomial coefficient is 6. so I need these 6 possible sets

A=[1 1 1 0 0 0 0;

1 1 0 0 0 1 0;

1 1 0 0 0 0 1;

0 1 1 0 0 1 0;

0 1 1 0 0 0 1;

0 1 0 0 0 1 1];

[Stephen23]

Why is

0 1 0 0 0 1 1

included, but

0 0 1 0 0 1 1
1 0 0 0 0 1 1

are excluded?

Although you write in your question that you want "all combinations", it seems that you actually want a subset of the combinations, but so far the rules to derive that subset are not clear to me.

[subhashree priyadarshini]

Sir for my problem I need the second element of each row or column must be 1. Thats why I want to exclude the following rows...

0 0 1 0 0 1 1;

1 0 0 0 0 1 1;

1 0 1 0 0 0 1;

1 0 1 0 0 1 0;

Answer 1

[Stephen23]

Brute force, not particularly efficient:

>> A = [1,1,1,0,0,1,1];
>> P = unique(bsxfun(@and,A,unique(perms(A),'rows')),'rows');
>> P(sum(P,2)~=3,:)=[]
P =
  0  0  1  0  0  1  1
  0  1  0  0  0  1  1
  0  1  1  0  0  0  1
  0  1  1  0  0  1  0
  1  0  0  0  0  1  1
  1  0  1  0  0  0  1
  1  0  1  0  0  1  0
  1  1  0  0  0  0  1
  1  1  0  0  0  1  0
  1  1  1  0  0  0  0

[subhashree priyadarshini]

Thank you so much sir.

Sir can you please tell me how to exclude the following rows from P matrix.

0 0 1 0 0 1 1;

1 0 0 0 0 1 1;

1 0 1 0 0 0 1;

1 0 1 0 0 1 0;

[Stephen23]

P(P(:,2)~=1,:) = []

[subhashree priyadarshini]

Thank u so much sir.