Finding the numbers having a particular factor,an upper limit and a specific set of digits. Question : 
How many numbers less than $50000$ can be formed which are multiples $6$ using the digits $0,1,2,3,4,5$?
My attempt : 
First of all, for a number to be a multiple of $6$, it should be a multiple of $2$ as well as $3$. So, the number must be even as well as it's sum of digits should be a multiple of $3$. Next, the number should have only the digits $1,2,3,4$ in the first position as it has to be less that $50000$. But, I face a problem, If I put $2$ or $4$ in the first position, then I will have one less even digit for the last position and there would be cases where I only use the digits $0,1,2,4,5$ or $1,2,3,4,5$ so as to make the number divisible by $3$. So, is there an efficient way to cover all the cases and get the answer quickly?
All help will be appreciated.

We put it in three cases as the last digit must be either $0;2$ or $4.$ (I assume that you can use each digit only once).
First case: The number is in the form $\overline{abcd0}$, which means the numbers $a,b,c,d$ are $4$ of the remaining numbers available $(1;2;3;4;5)$.
There are $5$ ways of picking four numbers from this set, we add $a,b,c,d$ along with $e$ for each pick:
$1+2+3+4+0=10$, not divisible by $3$.
$1+2+3+5+0=11$, not divisible by $3$.
$1+2+4+5+0=12$, divisible by $3$.
$1+3+4+5+0=13$, not divisible by $3$.
$2+3+4+5+0=14$, not divisible by $3$.
We conclude that $(a,b,c,d)$ must be a permutation of $(1;2;4;5)$ that satisfy $a<5$.
For each sub-case $a=1;a=2;a=4$, note that $(b,c,d)$ is a permutation of $3$ remaining numbers in the set, so each sub-case will have $3!=6$ possible outcomes for $b,c,d$.
In total, this first case  (including $3$ sub-cases) should have $6 \times 3=18$ satisfied numbers.
Second case: The number is in the form $\overline{abcd2}$, which means the numbers $a,b,c,d$ are $4$ of the remaining numbers available $(0;1;3;4;5)$.
There are $5$ ways of picking four numbers from this set, we add $a,b,c,d$ along with $e$ for each pick:
$0+1+3+4+2=10$, not divisible by $3$.
$0+1+3+5+2=11$, not divisible by $3$.
$0+1+4+5+2=12$, divisible by $3$.
$0+3+4+5+2=14$, not divisible by $3$.
$1+3+4+5+2=15$, divisible by $3$.
We conclude that $(a,b,c,d)$ must be a permutation of $(0;1;4;5)$ or $(1;3;4;5)$ that satisfy $a<5$ and $a\ne 0$.


*

*Case 2.1: $(a,b,c,d)$ is a permutation of $(0;1;4;5)$ that satisfy $a<5$ and $a\ne 0$. There are two sub-cases for this one ($a=1;a=4$), plus each sub-case has $6$ possible outcomes for $b,c,d$, so there are $6 \times 2 =12$ satisfied numbers.

*Case 2.2: $(a,b,c,d)$ is a permutation of $(1;3;4;5)$ that satisfy $a<5$ and $a\ne 0$. There are three sub-cases for this one ($a=1;a=3;a=4$), plus each sub-case has $6$ possible outcomes for $b,c,d$, so there are $6 \times 3 =18$ satisfied numbers.
So the second case should have a total of $12+18=30$ satisfied numbers.
Third case: The number is in the form $\overline{abcd4}$, this is similar to the second case, you can do it by yourself, after finishing it you can check the answer below.

 There are $78$ five-digit numbers less than $50000$ divisible by $6$ and can be formed by using the digits $(0;1;2;3;4;5)$

I think you should do similarly for the case one, two, three, four digit numbers, so here are two extra hints to do them:

 The number of ways to choose $k$ objects from $n$ objects, which the order does not matter, is: $$\frac{n!}{k!(n-k)!}$$
 For this problem the order of choosing numbers is important, so the number of permutations of $n$ distinct objects is $n!$.

I don't think this is a quick way, but it will surely help you to get the correct answer.
