ans : ( c ) n-r
explanation :
As rank is r, which is lesser than n, there should be infinite solutions! but wait,
A matrix represents linear transformation of a vector space. when the rank is less than column space, it simply means that, after applying that transformation, all vectors in n dimensions are mapped to vectors in r dimensions, (for example from 3d to 2d).
So, in this case, n dimension space got shrunk into r dimensions, that means n-r dimensions got vanished. a vector in any one of that n-r dimensions cannot be expressed as a linear combination of other dimensions (they are independent) but! all the infinite vectors in any of the n-r dimensions can be expressed as a linear combinations of other vectors in the same dimension.
therefore, there are n-r independent solutions
PS : If you think that you have no clue of what i am telling you, what you need is a geometrical understanding of linear algebra! if you are interested, watch https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab
Simple example :
consider the matrix
1 0 0
0 0 0
0 0 0
(shrinks 3D space to 1D)
now the independent solution sets are
0
y
0
(all vectors in y axis, put any thing to y ) and
0
0
z
(all the vectors in z axis, put any thing to z)
The infinite solutions which everyone says is all the vectors that lied in yz plane before applying this transformation!
that is
0
x
y
(that's your infinite solution set :-)