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So I know you could do this the tedious way of:

$(nCk) + (nCk_{+1}) + ... + (nCk_{=n})$

or the somewhat less tedious, but still not ideal:

$2^n - [(nC0) + (nC1) + ... + (nCk_{-1})]$

However, this problem to me is just screaming to think of it in terms of an $n$-bit integer, where a $1$ represents an element being present, a $0$ not present. so I guess what im asking is:

How would you determine the number of $n$-bit integers that contain at least $k$ $1$s?

Forgive me if my formatting is off, this is my first post. Tips appreciated!

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This problem is actually notoriously tough, and I do not believe it has a closed form solution. Note that, by symmetry, this is equivalent to summing over the first $k$ binomial coefficients, and then see here: https://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n

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  • $\begingroup$ Haha. Well now I feel silly. New to discrete maths, but have found even these elementary counting techniques quite interesting. Thanks for the response! $\endgroup$ – avcJav Dec 20 '15 at 6:07

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