Binary matrix for all possible unique results of X(i) variables

How to generate a binary matrix for all possible permutations of 'i' variables X, where " i " can be any number between 1 and infinite. I know that the resultant matrix will have 2^ i unique answers, but don't know how to generate it.  Let me explain by example:

variables x1, x2 (i=2) each with a possible value of 1 or 0, so the resultant matrx would be:

X2  X1
0     0
0     1
1     0
1     1

but what if i=120 for example, how do we generate the matrix for x(i) from 1 to 120? There would be 2^ 120 results.

The logic is simple, but I can not seem to translate to code:

x4 x3 x2 x1
0    0   0   0
0    0   0   1
0    0   1   0
0    0  1    1
0   1   0   0
0   1   0    1
0    1  1   0
0   1   1   1
1   0   0   0
1    0   0   1
1    0   1   0
1    0  1    1
1   1   0   0
1   1   0    1
1    1  1   0
1   1   1   1

generate matrix 2^i by i size
start with all x(i) values at 0
get value of x(i)
  if x(i)=1
      increment i
  else set x(i)=1 then start checking values from beginning
if i reaches count of total variables for that row OR leftmost variable is 1 move to next row

or something like above
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I don't know the mathlab programming language but I can give it to you in vbscript.  If you stick it in a file, say bitpat.vbs and run it from the command line under windows using

cscript bitpat.vbs

You'll get your patterns
bits = 5 ' number of bits required
count = 5^2
for c = 0 to count
   disp = ""
   val = c
   for b = 1 to bits
      if (val and 1) = 1 then
         disp = " 1" & disp
         disp = " 0" & disp
      end if
      val = val \ 2 ' shift right
   WScript.echo disp

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BrainstormerAuthor Commented:
cup, thanks for the VB script. It works for i=5, but I tested it for 6, or 7 and it fell short and stopped before all final values were 1. It also uses a string logic that does not truly "read" the previous values, which is OK, but not the "smart" solution I am looking for.

PS: I refuse to believe no one has ever had to come up with this problem before, and a true solution must exist. Come one programming experts, take a swing!
BrainstormerAuthor Commented:
I was able to resolve this using a MatLab function available here:

I also attached it for review purposes. The function code was hard for me to understand, I would still be interested into a more logical approach to this that's more language independent and easier to understand.
Sorry - the 5^2 should have been

count = (2^ bits) - 1

As for your program,  the notation is beyond me.
BrainstormerAuthor Commented:
that function is beyond me as well, but I only need to call it to get the desired results.

CUP, your updated script is working quite well, I will try to get a MatLab version of it up.

%combn.m function required
%define the matrix columns
%create the matrix M for all binary combinations
M=combn([0 1],i);
%display M on screen

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