The other tutor gave the probability of selecting the
first 8 correctly.  But that's not what was asked. 
You were asked for the probability of this:
After putting 18 names with 18 trees, discovering that
you have placed exactly 8 names correctly with the trees 
and placed all remaining 10 names incorrectly with trees.

You'd have to be studying combinatorial analysis to have 
such a problem as this.  But maybe you are.  If so, then:

We can choose the 8 trees to get correct in C(18,8) ways.  
However to get the number of ways to get all the other 10
incorrect, we must make use of the formula for the number 
of "derangements".  That is, the number of permutations 
of a1,a2,...an such that no ak is in position k. That 
number is !n or "n subfactorial", where   

!n = ⎣(e+e-1)n!]-⎣e·n!⎦  for n ≥ 2

where ⎣ ⎦ indicates the floor function. 

!10 = ⎣(e+e-1)10!]-⎣e·10!⎦ = 1334961

Number of successful ways = C(18,8)×(!10) = 43758×1334961=58415223438

Number of possible ways = 18! = 6402373705728000

So the probability of placing exactly 8 correct and exactly 10
incorrect is 58415223438/6402373705728000 = 0.0000091239946499,
approximately. 

Number of unsuccessful ways = 6402373705728000-58415223438 = 

6402315290504562

Odds in favor = 58415223438:6402315290504562

Odds against = 6402315290504562:58415223438

or odds against getting exactly 8 right and 10 wrong are about 
109600 to 1.

Edwin