You can do that but you have incorrectly calculated the probability that all of the missing stamps are unused.
You have 24 stamps, 10 of which are 'unused'. The probability the first stamp missing is unused is 10/24= 5/12. There are then 23 stamps, 9 of which are unused. The probability the second stamp missing is unused is 9/23. There are then 22 stamps, 8 of which are unused. The probability the third stamp missing is unused is 8/22= 4/11. The probability all three missing stamps are unused is $\displaystyle \frac{5}{12}\frac{9}{23}\frac{4}{11}= \frac{6}{101}$ so the probability there is at least one used stamp among the missing stamps is $\displaystyle 1- \frac{6}{101}= \frac{95}{101}$.
$\displaystyle P(C) = 1-\dfrac{14\cdot 13\cdot 12}{24\cdot 23\cdot 22} = \dfrac{415}{506}$.
$\displaystyle P(C|A) = 1-\dfrac{10\cdot 9\cdot 8}{20\cdot 19\cdot 18} = \dfrac{17}{19}$
If no green stamps are missing, then there are only 10 possible unused stamps that can be missing. So, the probability that at least one stamp is used is the complement of the probability that all three stamps are unused (none green). Essentially, you pretend that green stamps don't exist.