What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$?
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$
Looking at the prime numbers under $12$, we see that there are $\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{2^2}\right\rfloor+\left\lfloor\frac{12}{2^3}\right\rfloor=6+3+1=10$ factors of $2$, $\left\lfloor\frac{12}{3}\right\rfloor+\left\lfloor\frac{12}{3^2}\right\rfloor=4+1=5$ factors of $3$, and $\left\lfloor\frac{12}{5}\right\rfloor=2$ factors of $5$. All greater primes are represented once or none in $12!$, so they cannot be part of the square. Since we are looking for a perfect square, the exponents on its prime factors must be even, so we can only use $4$ of the $5$ factors of $3$. Therefore, the prime factorization of the square is $2^{10}\cdot3^4\cdot5^2$. To find the square root of this, we halve the exponents, leaving $2^5\cdot3^2\cdot5$. The sum of the exponents is $\boxed{\textbf{(C) }8}$