Question: $$ A \times B \times C \times D \times E \times F = 7 \times 10^7 $$

How can I find the number of ordered solutions for integers (I mean for integers $A,B,C,D,E,F$) so that they can satisfy the above equation?

Answer: 
The factorization of $7 \times 10^7 $ is simply $2^7 \times 5^7 \times 7 $.
Suppose for now we restrict the search for positive integer solutions only.
For non-negative integers $x\_i, y\_i, z\_i $, let $A = 2^{x\_1} \cdot 5^{y\_1} \cdot 7^{z\_1}, B = 2^{x\_2} \cdot 5^{y\_2} \cdot 7^{z\_2} , \ldots, F = 2^{x\_6} \cdot 5^{y\_6} \cdot 7^{z\_6} $.

Then their product is $2^7 \times 5^7 \times 7^1 = 2^{\sum\_{j=1}^6 x\_j} \times 5^{\sum\_{j=1}^6 x\_j} \times 7^{\sum\_{j=1}^6 x\_j} $.

$ \Rightarrow \displaystyle \sum\_{j=1}^6 x\_j = 7, \sum\_{j=1}^6 y\_j = 7, \sum\_{j=1}^6 z\_j = 1 $.

By Stars and Bars, like TravisJ pointed out, we have $ {7+6 - 1 \choose 6-1} = {12 \choose 5} $ ordered solution for $<x\_i>$ and $<y\_i>$, and ${1+6 - 1 \choose 6-1} = 6 $ ordered solution for $<z\_i>$.

Multiplying them together gives a total of $ {12 \choose 5} ^2 \times 6 = 3763584 $ ordered solution for positive integers $A$ to $F$.

If we drop the constraint for positive integers, then we could have:

Case 1: 6 Positive integers

Case 2: 4 positive integers, 2 negative integers

Case 3: 2 positive integers, 4 negative integers

Case 4: 6 Negative integers

Combining all these cases together yields a total of
$3763584 \times \left [ {6 \choose 0} + {6 \choose 2} + {6 \choose 4} + {6 \choose 6} \right ] = 120434688 $ ordered solutions.
This is essentially the same as: [How many ways are there to express a number as the product of groups of three of its factors?](https://math.stackexchange.com/questions/1273623/how-many-ways-are-there-to-express-a-number-as-the-product-of-groups-of-three-of/1273628#1273628) except instead of needing 3 distinct numbers you need 6.

First, factor $7\times 10^{7}=2^{7}5^{7}7^{1}$. Then apply the idea there. Keep in mind that your stars and bars method has a $5$ bars instead of $2$ in the solution. I'll let you read the post to see why it is true, but the final answer will be $\binom{7+5}{5}\binom{7+5}{5}\binom{1+5}{5}$.