A 10 digit positive number and ordered triplets A 10 digit positive number is said to be a “LearnHub”number if its digits are all distinct and it is a multiple of 11111. How many “LearnHub” numbers are there?

Find the number of ordered triplets (a, b, c) of positive integers for which LCM (a, b) =1000, LCM (b, c) = 2000 and LCM (c, a) = 2000.

How do i do this ? I have worked a lot on them.
Note that $1000=2^35^3$ and $2000=2^45^3$. What does this tell you about the prime factors of $a$, $b$, and $c$? Do you know what the LCM is of, say, $p^rq^s$ and $p^tq^u$, if $p$ and $q$ are primes? 

Can you see why any number with all 10 digits must be a multiple of 9? Can you see why any number that's a multiple of 9 and a multiple of 11111 must be a multiple of 99999? Now, $99999=100000-1$, so if you mutiply it by the 5-digit number abcde, you get abcde00000-abcde. From this you should be able to work out the possibilities for abcde to make the result contain all 10 digits. 

EDIT: Of course, there's always the computer programmer's solution, where you just ask your computer to look at all the 10-digit multiples of 11111 and count up how many have all 10 digits.
$1000 = 2^3 * 5^3$

$2000 = 2^4 * 5^3$

Since LCM of $a$ and $b$ is $1000$, they at the most contain $2$ three times and $5$ three times.

Also the LCM of terms containing $C$ is $2000$, the term $c$ should contain $2$ four times, as $a$ and $b$ can contain $2$ only three times.

so $c$ should either be

$2^4 = 16$

$2^4 * 5^1 = 80$

$2^4 * 5^2 = 400$

or $2^4 * 5^3 = 2000$. 

Either $a$ or $b$ should have $2^3$. Hence fixing one as $2^3$ other can take values from $2^0$ to $2^3$.
Hence total combination= $2*4 - 1$ (the one combination is $2^3,2^3$ which we took twice)$= 7$

2 of the three should have power of $5$ as $5^3$, whereas the third can take 4 values in terms of powers of $5$ $(5^0$ to $5^3)$. Hence total no of such arrangement = $(3!/2! *4) -2$(this is for $5^3,5^3,5^3$ which we took thrice) = $12 -2 =10$

Hence ordered triplets will be $10*7 = 70$