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!