A positive integer divisor of $12!$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23$
The prime factorization of $12!$ is $2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11$. 
This yields a total of $11 \cdot 6 \cdot 3 \cdot 2 \cdot 2$ divisors of $12!.$
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that the divisor can't have any factors of $7$ and $11$ in the prime factorization because there is only one of each in $12!.$ Thus, there are $6 \cdot 3 \cdot 2$ perfect squares. (For $2$, you can have $0$, $2$, $4$, $6$, $8$, or $10$ $2$s, etc.)
The probability that the divisor chosen is a perfect square is \[\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m\ +\ n = 1\ +\ 22 = \boxed{\textbf{(E) } 23 }\]
~mshell214, edited by Rzhpamath