There is 10 people that pick-up random number between 1 to 20 There is 10 people that pick-up random number between $1$ to $20$. More then one person can pick up same number (i.e. the pick-ups are independent).
What is the probability that the minimum number of all the people is grater then $8$?
My answer is $\frac{12^{10}}{20^{10}}$, I'm right?
And where I confused is about "What is the probability that the minimum number will be exactly $8$?  I don't know which is right:
$$\frac{10\cdot 13^9}{20^{10}}\;\text{or}\;\left(\frac{13}{20}\right)^{10}-\left(\frac{12}{20}\right)^{10}$$ 
Here is the conflict:
$\frac{10\cdot 13^9}{20^{10}}$ - One person must choose $8$ so we have 10 options. then, all the others can choose a number that is $8$ or bigger: $13^9$ [from 8 to 20] and we divide it by $\Omega$.
$\left(\frac{13}{20}\right)^{10}-\left(\frac{12}{20}\right)^{10}$ - is the number of people that can pick-up a number that is equal or greater then $8$ minus the number of people the can pick up a number that is grater then $8$.
Please help to understand which is the right one...
Thank you!!

Your answer $\dfrac{12^{10}}{20^{10}}$ to the first question is right: all $10$ people must pick numbers from among the $12$ numbers greater than $8$, the probability of which is $\left(\dfrac{12}{20}\right)^{10}$.
For the second question (minimum exactly $8$), your second answer $\left(\dfrac{13}{20}\right)^{10} - \left(\dfrac{12}{20}\right)^{10}$ is right, for the reason you stated: it's the probability that the minimum is greater than $7$, minus the probability that the minimum is greater than $8$.
In your other approach, when there are multiple people with the same choice $8$ (and the minimum is still exactly $8$), you count the same possibility multiple times for each choice of which of them you focus on as the "special" person with the minimum. You can take that into account as follows: suppose the number of people who chose exactly $8$ is $k$, which means that the remaining $10-k$ people chose numbers greater than $8$. The probability of this is 
$$\binom{10}{k} \left(\frac{1}{20}\right)^{k} \left(\frac{12}{20}\right)^{10-k}$$
As all $k$ such that $1 \le k \le 10$ are possible, the probability is got by adding them up, which gives, via the binomial theorem 
$$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k},$$ 
the same answer 
$$\left(\frac{1}{20} + \frac{12}{20}\right)^{10} - \binom{10}{0}\left(\frac{1}{20}\right)^{0} \left(\frac{12}{20}\right)^{10}$$
$$ = \left(\dfrac{13}{20}\right)^{10} - \left(\dfrac{12}{20}\right)^{10}$$
as before.
