Having more than two 6s with 10 dices 
So, the question is:
  Calculate the probability that 10 dice give more than 2 6s.

I've calculated that the probability for throwing 3 6s is 1/216.
And by that logic: 1/216 + 1/216 + .. + 1/216 = 10/216.
But I've been told that this isn't the proper way set it up.
Anyone having a good way to calculate this?
 A: The probability no 6s are thrown: $(\frac{5}{6})^{10}$.
The probability exactly 1 6 is thrown: $\binom{10}{1}(\frac{1}{6})^1 (\frac{5}{6})^9$
EDIT: $\binom{10}{2}(\frac{1}{6})^2 (\frac{5}{6})^8$
A: The easiest way to calculate this is to note that $ P(\#6s \le 2) + P(\#6s \gt 2) = 1$ That is that there must either be two or less or more than two sixes. So we can calculate it as $1-P(no\ 6s) - P(one\ 6) - P(two\ 6s)$. 
To get no sixes, we have to get the other 5 results on all other dice. To get one six we'll get a 6 on one die, but a different result on the other dice. Likewise we'll have to get two 6s, then 8 of the others.
This step is probably what you're getting confused about. We have to multiply these by the number of ways to 'pick' the die that gets the 6. So we'll be multiplying by $10\choose{n}$ for n 6s.
So the total sum is going to be $1 - {10\choose{0}}(5/6)^{10} - {10\choose{1}}(1/6)(5/6)^{9} - {10\choose{2}}(1/6)^2(5/6)^{8} \approx 0.22$
$10\choose 0$ is just for example, we're picking 0 dice so it's just 1
