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**1. Homework Statement**

choose x = (x1, x2, x3, x4) in R^4. It has 24 rearrangements like (x2, x1, x3, x4) and (x4, x3, x1, x2). Those 24 vectors including x itself span a subspace S. Find specific vectors x so that dimS is 0, 1, 3, 4

**3. The Attempt at a Solution**

So, i thought of it this way: 24 possiblities to arrange 4 numbers:

4*3*2*1 = 4! = 24

for dim S = 0 i got vector x = 0-vector;

for dimS = 1 i got x = (1, 1, 1, 1) since whichever way you rearrange it, it's still same;

for dim S = 4 i got x = (0, 0, 0, 1) or complement of this, since 1 can be carried through 4 times and new Lin indep vectors will result, which would be a basis and so dimS = 4;

but for dimS = 3 i can't figure it out, I tried all possible combinations of (1, 1, 0, 0) and (1, 1, 1, 0), but it does not work out....

it does not say that elements should be 0 or 1 necessarily, but i thought it was simple enough.....

Thanks in advance.