Suppose 40 distinguishable balls are to be distributed into 4 different boxes such that each box gets exactly 10 balls. Out of these 40 balls, 10 are defective and 30 are non-defective. In how many ways can the balls be distributed such that all the defective balls go to the first two boxes?\\
(A) $\frac{40!}{(10!)^4}$\\
(B) $\frac{30! \cdot 20!}{(10!)^5}$\\
(C) $\frac{20! \cdot 20!}{(10!)^5}$\\
(D) $\frac{30! \cdot 10!}{(10!)^4}$