NOT HW just practrice question
If eight distinct dice are rolled what is the probability that all six numbers appear ?
The chapter is INclusion and exclusion and the answer in the back of the text is :
$\frac{6^8}{6^8}-\binom{6}{1}\frac{5^8}{6^8}+\binom{6}{2}\frac{4^8}{6^8}-\binom{6}{3}\frac{3^8}{6^8}+\binom{6}{2}\frac{2^8}{6^8}-\binom{6}{1}\frac{1^8}{6^8}$
I am trying to understand this answer.. and what I follow is that the first term $\frac{6^8}{6^8}$ is the probability of the first number and then we subtract probability of the second number appearing $\binom{6}{1}\frac{5^8}{6^8}$ but why is it multiplied by $\binom{6}{1}$ or perhaps I am not interpreting the answer correctly. Could use some intuition n understanding this answer.
It's a binomial coefficient, given by $$ \binom{a}{b}=\dfrac{a!}{b!(a-b)!} $$
for $a,b\in\mathbb{N}$, $a\geq b$. Find more in http://en.wikipedia.org/wiki/Binomial_coefficient