Roll a fair dice 8 times, how many outcomes are there such that every side occurred at least once? I tried inclusion-exclusion.
There are $6^{8}$ possible outcomes, subtract off those with only 5 outcomes, add those with 4 outcomes... There are $\binom{6}{5}$ ways to choose those with 5 outcomes, $\binom{6}{4}$ to choose those with 4 outcomes...
so the answer is 
$\binom{6}{1}^8 - \binom{6}{5}\binom{5}{1}^8+\binom{6}{4}\binom{4}{1}^8-...$
Can someone tell me why this is wrong? 
 A: Use inclusion/exclusion principle:


*

*Include the number of outcomes with at most $6$ values, which is $\binom66\cdot6^8$

*Exclude the number of outcomes with at most $5$ values, which is $\binom65\cdot5^8$

*Include the number of outcomes with at most $4$ values, which is $\binom64\cdot4^8$

*Exclude the number of outcomes with at most $3$ values, which is $\binom63\cdot3^8$

*Include the number of outcomes with at most $2$ values, which is $\binom62\cdot2^8$

*Exclude the number of outcomes with at most $1$ values, which is $\binom61\cdot1^8$


So the total number of desired outcomes is:
$$\binom66\cdot6^8-\binom65\cdot5^8+\binom64\cdot4^8-\binom63\cdot3^8+\binom62\cdot2^8-\binom61\cdot1^8=191520$$
A: Here's a different approach to it. So the sets that we are allowed are of the form
$$1\,2\,3\,4\,5\,6\,a\,b$$
where $a$ and $b$ are any dice value. Now let's make combinations and permute them. If $a=b$, then we have a string of 8 numbers with a digit repeated three times, which has a total number of permutations $\frac{8!}{3!}$. If $a\neq b$, then we have a string of 8 numbers with two digits repeated twice each, which has a total number of permutations $\frac{8!}{2!2!}$. There are 6 combinations such that $a=b$, and 15 combinations such that $a\neq b$. Putting this information together, the number of permutations such that each number occurs at least once is
$$6\cdot\frac{8!}{3!}+15\cdot\frac{8!}{2!\,2!}=191520$$ 
