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I'd like to find the sample size for which, in a combination with replacement, the probability of having at least one object of each k class is greater than $p$. Each object can take $k$ levels. I think (but I'm not sure) that for a sample size of $n$ and $k$ different classes, the number of different combination with at least one in each $k$ class is : $\sum_{i=1}^{n-k+1}{i}$.

Thus, I'd like to solve this equation : $\frac{\sum_{i=1}^{n-k+1}{i}}{\binom{k+n-1}{n}} = p$ or if I am right, this is $\frac{\frac{1}{2}(n-k+1)(n-k+2)}{\frac{(k+n-1)!}{n!.(k-1)!}} = p$

$p$ is a proportion which should be 0.95 or 0.99, and $k$ is known (256).

Unfortunately, i don't know how to handle either of these ! I read this post, but in my case, as $n$ is in both up and down, i'm lost !

Thanks for any hint on this and sorry for the mistake I introduced the first time !

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jtextori
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  • Is $p$ a prime? Are you wanting to solve for $n$ and $k$? Are one of these variables fixed at a particular number? This could be taken several ways, I think. – Clayton Jul 17 '13 at 19:36
  • p is a proportion, so $0 < p < 1$. I should have tell this ! And no, I forgot also to tell that I know $k$ and $p$. But in my case, $k$ is big (256) so even trying a few example of n is not possible unless I simplified this. – jtextori Jul 17 '13 at 19:38
  • For nonnegative integers, $\binom{m}{n}$ is always a nonnegative integer, so $\binom{n+k-1}{n} = p$ for $0 < p < 1$ can only have a solution if $k < 1$, $k$ not an integer or $k < 1-n$. – Daniel Fischer Jul 17 '13 at 19:47
  • Well, sorry for that again, i wanted to simplified, i will edit the question. – jtextori Jul 17 '13 at 19:50

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