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I want to solve combination $C(n,k)$ for $k$ but don't know how. Suppose I have $n=10$ and $C(n,k)=15$. What will $k$ be?

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    There is no integer $k$ such that $C(10,k) = 15.$ Perhaps you should give more background on how you came to need to solve such an equation. – 2017-01-13
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    Maybe I just don't know the rules for solving factorials... Ok, for n=10 and k=2, C(n,k)=45, right? Then if k=? how to find it knowing 'n' and C(n,k)? – 2017-01-13
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    For small numbers, use trial and error. For very large numbers you might get an approximate answer using some form of Stirling's Formula. Neither of these really is a "rule for solving" of the kind you seem to be looking for. – 2017-01-13
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    And reasoning with divisibility pairs reasonably well with the trial and error. Here, for instance, we may assume $k \le 5$ by symmetry. But because $7$ doesn't divide $C(10, k)$, then $k \le 10 - 7 = 3$. Clearly $k \neq 0, 1$, leaving very few things to check. – 2017-01-13
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    Ok, why I need it? My proposal for solution may be wrong... Suppose I have 10 workers and 3 tasks. Every one must complete each task (and only once per worker), so the number of total 'completions' is 10 * 3 = 30. Now if the number of 'completions' is 15, how to decide the minimum (or maximum) number of workers who took part in the task completion process? – 2017-01-13
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    Maybe using Stirling formula for getting an approximation and then by trial and error see if by increasing or decreasing $k$ you get closer to $C(n,k)$. At some point you will either reach exactly $C(n,k)$ or will overpass the value which will mean there is not answer since $C(n,k)$ is strictly increasing for $k \leq n/2$ and symetric. – 2017-01-13
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    More generally, you could search for it using binary search, which is pretty fast as long as you can compute $C(n,k)$ quickly. But it seems you are only interested in small examples, if I'm correct, in which case 'naive' trial and error is definitely the way to go. – 2017-01-13

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