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!

There are 5 types of possibilities for the unordered exponents $a_1,\cdots,a_5$ for 2:
$\textbf{1)}$ $6+0+0+0+0,\;\; 2+1+1+1+1$  $\;\;$(4 of a kind)
$\textbf{2)}$ $5+1+0+0+0,\;\; 3+0+1+1+1,\;\;4+2+0+0+0$  $\;\;$(3 of a kind)
$\textbf{3)}$ $4+1+1+0+0,\;\; 0+2+2+1+1$  $\;\;$(2 pairs)
$\textbf{4)}$ $3+3+0+0+0,\;\; 0+0+2+2+2$  $\;\;$(full house)
$\textbf{5)}$ $3+2+1+0+0$  $\;\;$ (1 pair)

Since we have the same 5 possibilities for the unordered exponents $b_1,\cdots,b_5$ for 5, 
we can count the number of ways to pair up these types to get distinct divisors, using the fact that digits matched to repeated digits must be distinct, and their order doesn't matter:
$\textbf{1)}$ and 5) $\;$and$\;$ $\textbf{5)}$ and 1):  $\;\;\;2(2\cdot1)(1)={\color{red}4}$ ways
$\textbf{2)}$ and 2):  $\hspace{1.03 in}\;\;\;(3\cdot3)(1)={\color{red}9}$ ways
$\textbf{2)}$ and 3) $\;$and$\;$ $\textbf{3)}$ and 2):  $\;\;\;2(3\cdot2)(2)=\color{red}{24}$ ways
$\textbf{2)}$ and 5) $\;$and$\;$ $\textbf{5)}$ and 2):  $\;\;\;2(3\cdot1)(1+2\cdot3)=\color{red}{42}$ ways
$\textbf{3)}$ and 3):  $\hspace{1.03 in}\;\;\;(2\cdot2)(1+2\cdot2)=\color{red}{20}$ ways
$\textbf{3)}$ and 4) $\;$and$\;$ $\textbf{4)}$ and 3):  $\;\;\;2(2\cdot2)(1)=\color{red}{8}$ ways
$\textbf{3)}$ and 5) $\;$and$\;$ $\textbf{5)}$ and 3):  $\;\;\;2(2\cdot1)(2\cdot3+3\cdot2)=\color{red}{48}$ ways
$\textbf{4)}$ and 5) $\;$and$\;$ $\textbf{5)}$ and 4):  $\;\;\;2(2\cdot1)(3)=\color{red}{12}$ ways
$\textbf{5)}$ and 5):  $\hspace{1.03 in}\;\;\;3\cdot3+3\cdot3\cdot2=\color{red}{27}$ ways

This gives a total of $\color{red}{194}$ ways of expressing $10^6$ as a product of 5 distinct positive integers.
