Crazy Matrices

First of all we have to look at how many ways are there to form a row. We know that in a row there are $3$ elements and each element can be filled in $2$ ways.

Number of ways of forming a row = $2^3=8$ ways.

Of the $8$ rows we have to select 3 rows and the three rows can be arranged in $3!$ ways. Hence, total number of elements in S are $\dbinom{8}{3}3!$ = $336$

Let A be any binary matrix. Fix the first row in A ( say 1 0 0 ). We choose second row such that it is distinct from first row. It can be done in $2^{3}-1$ ways (except 1 0 0 ). Now fix any such second row ( say 1 0 1). We choose third row such that it is distinct from both the first and the second one. So there are $2^{3}-2$ ways (except 1 0 0 and 1 0 1). Now we can choose the first row itself in $2^{3}$ ways. So there are totally * $8 \times 7 \times 6$ = * 336 ways