1000 items, 10 defective, m selected randomly for inspection.
Find the value of m such that the probability of selecting at least 1 defective is .90.
$$p = 1 - \frac{{990}\choose{m}}{{1000}\choose{m}} = .90$$
i.e. 1 minus the probability of selecting no defective items.
Simplifying a bit (if it helps):
$$ \frac{{990}\choose{m}}{{1000}\choose{m}} = \frac{990!}{(990-m)!m!} \div \frac{1000!} {(1000-m)!m!} = \frac{990!}{(990-m)!m!} \times \frac{(1000-m)!m!} {1000!} $$
$$ = \frac{(1000-m)!}{(990-m)!} \times \frac{990!}{1000!} = \frac{(1000-m)!}{(990-m)!} \div \frac{1000!}{990!} = \frac{(1000-m)P10}{1000P10}. $$
So
$$1 - \frac{(1000-m)P10}{1000P10} = .90 \implies \frac{(1000-m)P10}{1000P10} = .10$$
Through trial and error I got $m = 205$ but I want to know if there's an algebraic way to find a solution for $m$.
For X ~ Binomial{$n = \space?, p = 0.01, k = 0$}, you need to find $$P(X = 0) = 0.1 \color{blue} \implies \binom{n}{0}(0.01)^0(0.99)^{n} = 0.1$$
Since the first first two terms are unity, and then taking logs: $$(0.99)^{n} = 0.1 \color{blue} \implies n = \frac{\text{log } 0.1}{\text{log } 0.99} \approx 229$$
Note that the answer I get is different from yours.