The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
Use properties of exponents to move the squares outside the brackets use difference of squares.
\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]
Using the binomial theorem, we can see that the number of terms is $\boxed{\mathrm{(B)}\ 5}$.