Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

Let there be two families with 3 members each (say Family A and Family B), and one family with 4 members (Family C).

Step 1: Since members of the same family must sit together, treat each family as a single block.
Thus, there are 3 blocks: A, B, and C.
They can be arranged in  $3! = 6$ ways.

Step 2: Now arrange members within each family:
Family A (3 members): $3!$ ways
Family B (3 members): $3!$ ways
Family C (4 members): $4!$ ways

Step 3: Total number of arrangements = $3! \times 3! \times 3! \times 4!$

Step 4: Simplify:
$3! = 6$ and $4! = 24$
$\Rightarrow (3!)^3\times4!$

∴ Total number of ways = $\boxed{5184}$