jmc

The prime factorization of a positive integer factor of 2160 is of the form $2^a\cdot 3^b\cdot 5^c$ where $0\le a\le 4$, $0\le b\le 3$, and $0\le c\le 1$. A positive integer is a perfect square if and only if all the exponents in its prime factorization are even. Therefore, we are free to choose $a$ from the set $\{0,2,4\}$ and $b$ from the set $\{0,2\}$. In total, we have $3\times 2=\boxed{6}$ choices for the exponents in the prime factorization of a perfect square factor of 2160.