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Out of N switches, what is the formula for the number of combinations with on switches >= K, assuming K <= N

It's basically

$$ \begin{pmatrix}{N \\ K}\end{pmatrix} + \begin{pmatrix}{N \\ K + 1}\end{pmatrix} + .... + \begin{pmatrix}{N \\ N-1}\end{pmatrix} + \begin{pmatrix}{N \\ N}\end{pmatrix} $$

but is there a more concise version?

I rewrote it as

$\sum\limits_{i=K}^N \begin{pmatrix}{N \\ i}\end{pmatrix}$

$\sum\limits_{i=K}^N \cfrac{N!}{i!(N-i)!}$

N! $\sum\limits_{i=K}^N \cfrac{1}{i!(N-i)!}$

I know there's going to be N - K + 1 terms in the sum, so then letting

j = i - K

I rewrote the sum as

N! $\sum\limits_{j=0}^{N-K} \cfrac{1}{(j+K)!(N-j-K)!}$

But then I'm out, can it be taken any further? Perhaps even to a formula without a summing symbol?

Matt
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  • Have you ever heard of hypergeometric series? I think the summation can be tackled by summing up the partial sum of hypergeometric with Gosper's algorithm. – robit Dec 03 '15 at 08:15
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    Duplicated. Please refer [link]http://math.stackexchange.com/questions/69532/partial-sum-of-rows-of-pascals-triangle – robit Dec 03 '15 at 08:38

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