A car park has 10 empty spaces. 5 cars park in the car park. What is the probability of no two cars being parked next to each other?

For a systematic approach:

Note that the possible patterns must have the form $$0^a\,X\,0^b\,X\,0^c\,X\,0^d\,X\,0^e\,X\,0^f$$

Where $0$ is an empty space, $X$ is a car, $a+b+c+d+e+f=5$ and $b,c,d,e>0$. Letting $b'=b-1,c'=c-1,d'=d-1,e'=e-1$ we see that $a+b'+c'+d'+e'+f=1$ so exactly one of these is non-zero (and that one is equal to $1$). That translates to $6$ options, confirming your solution.

Park $5$ cars on $6$ spots -- in $6=\binom{6}{5}$ ways. Add between each car one extra spot and get a proper parking on $6+4=10$ spots.