Perfect square divisors of $12!$

The prime factorization of $12!$ is $2^{10} * 3^{5} * 5^{2} * 7 * 11$ .

Thus any perfect square divisor will be of the form $2^{a} * 3^{b} * 5^{c}$ , with

$6$ choices for $a$ , (namely $0, 2, 4, 6, 8, 10$ ),

$3$ choices for $b$ , (namely $0, 2, 4$ ), and

$2$ choices for $c$ , (namely $0, 2$ ).

Thus by the product rule $12!$ has a total of $6 * 3 * 2 = \boxed{36}$ perfect square divisors.

The prime factorization of 12! is $2^{10}$ X $3^{5}$ X $5^{2}$ X $7^{1}$ X $11^{1}$ .

We can write the former equation as:

( $4^{5}$ X $9^{2}$ X $3^{1}$ X $25^{1}$ X $7^{1}$ X $11^{1}$ ) .

Thus, convert the prime factors powers into square number powers.

Now, no. of Square divisors will be (5+1) X (2+1 ) X (1+1)

(add 1 to the powers of each divisors)

Therefore total no. of factors are 6 X 3 X 2 = $\boxed{36}$