choosing the right value to calculate a probability A lot of $n$ itms contains $k$ defectives, and $m$ are selected randomly and inspected. How should the value of $m$ be chosen so that the probability that at least one defective item turns up is 0.90?
Attempt:
So thinking about it i figured it would be easier to figure out the complement of this thus $1 - p(A^c)$ where A = event.
As such:  $$1 - \frac {{n-k\choose m}{k\choose 0}}{{n\choose m}} $$
But now i am having an issue in getting rid of some of the factorials. Suggestions? Did i interpret it right?
 A: An analytic  solution  to the exact hypergeometric
seems messy.
Some simplifications is possible if you express
the binomial coefficients in terms of factorials: to begin,
two factors $m!$ cancel. You might be able to use Sterling's formula for an approximation to the rest, but that formula works best for large values, and I'm not sure how accurate the result would be in your case.
The exact quantity you show in your question is $P(X = 0),$ where $X$ has a hypergeometric distribution based selecting $m$ items without replacement from a population of $n$ items of which $k$ are defective. Using $p = k/n,$, the mean of this distribution is $\mu = mp,$ and standard deviation
$$\sigma = \sqrt{mp(1-p)\frac{n-m}{n-1}}.$$
For $m$ sufficiently large and $p$ not too close to 0 or 1,
$X$ has approximately a normal distribution with mean $\mu$
and SD $\sigma.$ So we seek $m$ just large enough that
$$P(X = 0) = P(X \le 0.5) \approx P(Z \le (0.5 - \mu)/\sigma) = .9.$$
Thus we could solve $(0.5 - \mu)/\sigma = 1.28$ for $m$ in terms of $k$ and $n$ when $\mu$ and $\sigma$ are expressed in terms of $m, n,$ and $k.$
Example: Letting $n = 100$ and $k = 5$ we use computations in R to find the exact value $m = 37$ from the hyergeometric distribution and the approximate value $m = 38$ from the normal approximation. We also illustrate that a binomial model with $m$ trials and probability $p$ of a defective on each draw (sampling with replacement) does not give a good value of $m$ in this example. 
 n = 100;  k = 5;  p = k/n;  m=1:(n-1)
 mu = m*p;  sd = sqrt(m*p*(1-p)*(n-m)/(n-1))
 # Normal approximation
 nor = 1 - pnorm(0.5, mu, sd)
 min(m[nor >= .9])
 ## 38

 # Exact hypergeometric computation
 hyp = 1 - dhyper(0, k, n-5, m)
 min(m[hyp >= .9])
 ## 37

 # Binomial approximation (unsatisfactory for this n and k)
 bin = 1 - dbinom(0, m, k/n)
 min(m[bin >= .9])
 ## 45

When the lot size is large enough that $m$ turns out to be very much smaller than $n,$ then the binomial works well. For example,
with $n = 1000$ and $k=200,$ the exact hypergeometric value
is $m = 13$ and both the normal and the binomial approximations
give $m = 11.$
The binomial approximation has the advantage
that it is very easy to solve $(1 - k/n)^m \approx 0.1$ for $m$ in terms of
$k$ and $n.$
