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 $194$.
I think the analogy with the permutations of letters is making this problem more complicated than it needs to be.
Using the restriction that the number has at least one seven, you can first find the numbers that have exactly one $7$, then the numbers that have two $7$s, and then the number that has three $7$s and then add the results.
To find the number of numbers, think of choosing a digit for each spot: _ _ _
For one seven, you can fix a $7$ in a spot, say the first one, so the number looks like 7_ _ and note that for the other spots you can have any of the other 9 digits ($0,1,2,3,4,5,6,8,$ or $9$), so there are $9\cdot 9=81$ such numbers.
For numbers of the forms _ 7 _ and _ _ 7 the count is different because the first digit cannot be $0$, so there are $8\cdot 9=72$ possibilities for each.
Thus, in total there are $72+72+81=225$ three-digit positive integers with one seven as a digit.
Two sevens: For the form _77 there are 8 possibilities because the first spot cannot be 0, and for each of the forms 7_7 and _ _7 there are 9 possibilities, so in total there are $8+9+9=26$ three-digit positive integers with one seven as a digit.
Three sevens: There is only one, $777$.
So in total there are $225+26+1=252$ three-digit integers with a seven as a digit.
Best Answer
Hint: Solve the problem for $abc = 10^6$. This has $ {8 \choose 2}^2=784$ solutions.
Count the number of solutions where $a=b=c$.
Count the number of solutions where $a=b$ or $b=c$ or $c=a$.
Count the number of solutions where $a, b, c$ are pairwise distinct.
Account for your repeated counting above, to find the cases where $a \leq b \leq c$.
Motivation: Bars and stripes, Orbit Stabilizier Theorem.