In how many ways can $1000000$ be expressed as a product of five distinct positive integers? I'm trying to solve the following problem:
"In how many ways can the number $1000000$ be expressed as a product of five distinct positive integers?"
Here is my attempt:
Since $1000000 = 2^6 \cdot 5^6$, each of its divisors has the form $2^a \cdot 5^b$, and a decomposition of 1000000 into a product of five factors has the form
$$
1000000 = (2^{a_1} \cdot 5^{b_1})(2^{a_2} \cdot 5^{b_2})(2^{a_3} \cdot 5^{b_3})(2^{a_4} \cdot 5^{b_4})(2^{a_5} \cdot 5^{b_5})
$$
where $a_i$ and $b_i$ are nonnegative integers which satisfy the
conditions
$$
a_1 + a_2 + a_3 + a_4 + a_5 = 6, b_1 + b_2 + b_3 + b_4 + b_5 = 6
$$
The total number of systems of $a_i$ which satisfy the first equation is $210$ and the same number is for $b_i$. Thus the total number of decompositions is $210 \cdot 210 = 44100$. 
However, in this enumeration, factorizations which differ only in the brder of the factors have been counted separately; that is, some factorizations are counted several times each. 
To get the number of distinct unordered decompositions I must, at first, substract the number of ordered decompositions with at least two identical factors and, at second, divide resulting number by $5!$ to leave only unordered ones.
And I'm stuck at the step of counting the number of ordered decompositions with at least two identical factors. 
The number of decompositions with $k$ identical factors is $(\lfloor\dfrac{a}{k}\rfloor + 1)(\lfloor\dfrac{b}{k}\rfloor+1){5 \choose k}$, that is, the number of decompositions with two identical factors is $16 \cdot {5 \choose 2} = 160$, three identical factors - $9 \cdot {5 \choose 3} = 90$, four ones - $4 \cdot {5 \choose 4} = 20$ and five ones - $4 \cdot {5 \choose 5} = 4$. Thus the number of distinct decompositions must be equal $\dfrac{44100-160-90-20-4}{5!}$, but this number is not integral. I suppose that the numbers of decompositions with $k$ identical factors overlap and I misuse inclusion-exclusion principle. But I have no idea how I can count that overlapped decompositions.
Please, help!
Thanks!

For $n{\ge}0$, $\;$ let $F(n)$ be the ways in which the number $10^n$ can be expressed as a product of five distinct positive integers.
Then
$$F(n)=\left(\frac{1}{5!}\right)\left(24\left(\left\lfloor\frac{n}{5}\right\rfloor-\left\lfloor\frac{n-1}{5}\right\rfloor\right)^{2}
+(-30)\left(\left\lfloor\frac{n}{4}\right\rfloor+1\right)^{2}
+(-20)\left(\left\lfloor\frac{n+2}{2}\right\rfloor-\left\lfloor\frac{n+2}{3}\right\rfloor\right)^{2}
+20\left(\left\lfloor\frac{(n+2)(n+3)}{6}\right\rfloor\right)^{2}
+15\left(\frac{2n^{2}+10n+11+(2n+5)(-1)^n}{16}\right)^{2} 
+(-10)\left(\frac{4n^{3}+30n^{2}+68n+45+3(-1)^{n}}{48}\right)^{2}
+\left(\frac{(n+1)(n+2)(n+3)(n+4)}{24}\right)^{2}
\right)$$
The answer for this question is $F(6)=194$.
$$F(10)=6355,\quad F(100)=175519523577,\quad  F(1000)=14758828112205481602.$$
