Number of permutations for hiring mr.X

Question

Let be $9$ candidates for a job hiring (mr.X included). There are $3$ judges. Every judge defines a priority list ($1st-9th$) for the candidates (1st being the best and 9th the worst). A candidate is being hired only if they exist in the first $3$ spots of each of the three lists. In how many cases is mr.X being hired?

I am thinking that for each judge there are :

$ (9-1)! * 3 $

permutations (cases) for mr.X to be hired. Given the above, then for all three judges it is:

$(8!*3)^3$

cases.

But that seems like extremely many cases.

Answer 1

Yes correct. The probability to be in the first 3 positions on 9 total is $\frac{3}{9}=\frac{1}{3}$

Then with 3 judges the probability is $(\frac{1}{3})^3$