We obtain all ordered pairs $(x,y)$ such that $xy=N$ and $\gcd(x,y)=1$ by choosing a subset of the set of prime divisors of $N$, and giving all these primes to $x$, and giving the rest of the primes to $y$. A set of $n$ elements has $2^n$ subsets, so there are $2^n$ ways to produce a suitable $x$, and hence $2^n$ ordered pairs $(x,y)$.
We now have a count of the ordered pairs. But we want to count the number of factorizations as a product of two relatively prime numbers, and for example $(20)(3)$ is considered to be the same factorization as $(3)(20)$. For $n\ge 1$, to count the unordered pairs, divide the number of ordered pairs by $2$. We get $2^{n-1}$. For a more familiar example, there are $(52)(51)$ ordered pairs of $2$ cards, but there are only $(52)(51)/2$ "hands" of two cards.
The only case where we do not divide by $2$ is the case $N=1$, that is, $n=0$, where the only factorization is $(1)(1)$. So the exact result is that the number of factorizations of the desired type is $2^{n-1}$ if $n\ge 1$ and $1$ if $n=0$.
Comment: Consider the product
$$(1+a+a^2+\cdots+a^p)(1+b+b^2+\cdots+b^q)(1+c+c^2+\cdots+c^r)\cdots \qquad(\ast)$$
Look at a typical full term $1+p+p^2+\cdots +p^k$. We produce the suitable divisors $x$ of $N$ by deciding whether we will use the $1$ or the $p^k$. No other choice is possible, because if we choose $p^i$ where $0<i<k$, then $N/x$ (in our notation, $y$) will be divisible by $p$, so $x$ and $y$ will not be relatively prime.
Thus at every term in the product $(\ast)$, we have $2$ choices, "None" or "All." There are $n$ terms in the product, so there are $2^n$ ways to produce $x$. Now argue as before that if $n \ge 1$, every suitable factorization is produced twice by this process.
The number in question is the coefficient of $x^6 y^6 z^5$ in the product
$$\prod_{0\le i,j\le 6}(1+x^i y^j z).$$
Here $z$ counts the number of factors (we want $5$); $x$ and $y$ tag the powers of $2$ and $5$ respectively; we want both of those to be $6$. Distinctness is ensured because, for each factor $2^i 5^j$, we either include it once (thereby multiplying by $x^i y^j z$) or we don't include it (and multiply by $1$).
Multiplying out the entire product is unwieldy, but you can compute it mod $(x^7,y^7)$, which eliminates a whole mess of terms you don't need. The coefficient of $x^6 y^6$ turns out to be
$$5 z^7+64 z^6+194 z^5+235 z^4+123 z^3+24 z^2+z,$$
which enumerates the number of ways to write $1000000$ as a product of $k$ distinct factors, for all values of $k$. Taking $k=5$ we confirm Christian's answer of $194$.
Best Answer
As you observed, $10800 = 2^4 \cdot 3^3 \cdot 5^2$.
Thus, each factor has the form $2^a3^b5^c$, where $0 \leq a \leq 4$, $0 \leq b \leq 3$, and $0 \leq c \leq 2$. If the two factors are $2^{a_1}3^{b_1}5^{c_1}$ and $2^{a_2}3^{b_2}5^{c_2}$, then \begin{align*} a_1 + a_2 & = 4 \tag{1}\\ b_1 + b_2 & = 3 \tag{2}\\ c_1 + c_2 & = 2 \tag{3} \end{align*} Equation 1 is an equation in the nonnegative integers with $5$ solutions, namely $(4, 0), (3, 1), (2, 2), (1, 3), (0, 4)$. Similarly, equation 2 is an equation in the nonnegative integers with $4$ solutions, and equation 3 is an equation in the nonnegative integers with $3$ solutions. Thus, there are $$5 \cdot 4 \cdot 3 = 60$$ ordered pairs of factors. Since $10,800$ is not a perfect square, each pair contains two distinct factors. Thus, we have counted each unordered pair of factors twice, so there are $$\frac{5 \cdot 4 \cdot 3}{2} = 30$$ unordered pairs of factors of $10,800$.