I don't know that there is a "proper way". There will definately be easier and harder ways.
But anyway, I would approach this by looking at the number of ways to choose 5 numbers with exactly two equal, then permuting these to all their possible orders.
$(x,x,a,b,c)$ is the set of numbers to permute. This only represents the set of numbers, it does not mean that we are only looking at the number $xxabc$. Later on, when we permute, this set will represent all possible ways of ordering $(x,x,a,b,c)$. for instance, $xxabc,axxbc,xbcax$ etc.The symbol $x$ represents the repeated number.
For the first case, we do not include zero because it can't go in the first position, which leaves us with only 9 numbers to choose from.
To fill out a, b and c in all the possible ways is $\binom{8}{3}$. We use 8 instead of 9 because we need to leave a number that $x$ can be. Then we multiply this by the number of ways to order these 5 digits, which is $5!$. Then we divide this by 2 because we don't want to count the two ways that $x$ can be ordered.
There are 9 different ways to choose which value $x$ will have (becasue we are neglecting zero at this point), so we have to multiply by 9.
So by this point we have:
$9*\binom{8}{3}*\frac{5!}{2}$
This covers numbers such as 13325, 66123, 79827 etc.
The next case is for when we have a single zero, which is slightly different because it can't go in the first position. In this case we have 10 numbers to choose from (0,1,...,9)
$(x,x,0,a,b)$ is the set we want to permute. The number of ways to fill out a and b are $\binom{8}{2}$ because we already have chosen two distinct digits. Now, out of the 5 digits, we can place only 4 in the first position (because zero can't go there), then we can place 4 in second, 3 in third etc, so instead of $5!$, in this case we get $4*4*3*2*1=4*4!$
We also have to divide by two because we don't want to count the number of ways that $x$ can be ordered. Then we multiply by 9 because again, the two $x$'s can be any number from 1 to 9.
This covers numbers such as 10233, 11098, 99054 etc.
By this point we have:
$9*\binom{8}{3}*\frac{5!}{2}+9*\binom{8}{2}*\frac{4*4!}{2}$
Now finally for the case when $0$ is the repeated number.
We will ignore the first digit for now. This leaves us with the four digit set $(0,0,a,b)$. There are $\binom{9}{2}$ ways to fill out a and b here. There are $4!$ ways to order these elements. We divide by 2 because we don't want to count the number of ways that 0 can be ordered. By this point, we have chosen 3 distinct numbers with 7 remaining to chose from for the first position. So to count for the different numbers that can take the first position, we multiply by 7.
This covers numbers such as: 10023, 98700, 20980 etc.
All together this gives a total of:
$\binom{9}{2}*\frac{4!}{2}*7+9*(\binom{8}{3}*\frac{5!}{2}+\binom{8}{2}*\frac{4*4!}{2})=45360$
Best Answer
Inclusion and exclusion is the way to do it. It's the number divisible by two of the factors, minus the number divisible by three of the factors, plus the number divisible by all four. As Matthew Daly said in a comment, you have to treat all the cases separately. For example, a number is divisible by all four factors if and only if it is divisible by their greatest common divisor, namely $420$. The number of multiples of $420$ less than or equal to $10000$ is $$\left\lfloor\frac{10000}{420}\right\rfloor=23$$
I leave the rest to you.