$7200$ can be written as $2^{5} * 3^{2} * 5^{2}$
Therefore, there are $(5+1) * ( 2+1) * (2+1) = 54$ divisors including $1$, and $7200$.
Hence, the total divisors (excluding $1$ and $7200$) = $54-2 = 52$
Here's some justification to support the above method :
Let $n\in\mathbb{N}$. Then by fundamental theorem of arithmetic we can write $n\in \mathbb{N}, n\neq 1$ by $n=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$ where $p_1,p_2,\dots p_k$ are prime and $a_k\in\mathbb{N}$, $k=1,2,...,k$. Hence, number of divisors of $n$ = $(a_1+1)(a_2+1)\cdots (a_k+1)$.
More on this here.
Answer ( C )
