In how many options can one cast 10 game cubes in different colors so that all the digits 1,2,3,4,5,6 will apear? I study discrete and I missed some lessons. Can you help?
The problem:
We have 10 game cubes, each in a different color. The question is what is the number of options to throw all the 10 cubes and get all the digits from 1 to 6. I thought the solution is the number of options that exists when throwing 10 different-color game cubes, minus the number of options in cases not all the digits are found, such as (1,2,3,4,5,1,2,3,4,5) or (1,1,1,1,1,1,1,1,1,1). 
Thanks a lot! :)
 A: This is an example of the inclusion-exclusion principle.  If you search the site you will find many applications.  We start with the $6^{10}$ possible rolls and subtract all the ones that are missing a number.  How many is that?  There are $6$ ways to select the missing number and $5^{10}$ ways to choose the numbers you do get.  Unfortunately, we have subtracted the ones missing two numbers twice each, so we need to add them back in once.  That is ${6 \choose 2}$ ways to choose the missing numbers and $4^{10}$ to choose the ones you have.  So far we have $6^{10}-{6\choose 1}5^{10}+{6 \choose 2}4^{10}$ and you have to worry about how many times we have counted the throws with only $1,2,3$ different numbers.
A: Use the principle of inclusion and exclusion. Say a throw has property $k$ if the $k$-face doesn't show up, you want the number of throws with none of the 6 properties.
A: This might be an overkill if you are beginning with discrete math. But I have answered a similar one before.
The number of options $a(k,m,n)$ is given by:
\begin{align*}
  a(k,m,n) = \binom{k}{m}\cdot \sum_{i=0}^{m} (-1)^i \, \binom{m}{i}\cdot (m-i)^n
\end{align*}
where 
$k$: no. of different faces on a die
$m$: exactly this no. of faces are shown up
$n$: no. of different colored dice are thrown
For your question, $k=6, m=6, n=10$, and hence:
$$a(6,6,10) = 16435440$$
