/me wonders if pythag is really a genius

by gauss1181, Apr 12, 2009, 7:47 PM

Let $ n$ and $ k$ be positive integers with $ k \geq n$ and $ k - n$ an even number. Let $ 2n$ lamps labelled $ 1$, $ 2$, ..., $ 2n$ be given, each of which can be either ''on'' or ''off''. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from off to on).

Let $ N$ be the number of such sequences consisting of $ k$ steps and resulting in the state where lamps $ 1$ through $ n$ are all on, and lamps $ n + 1$ through $ 2n$ are all off.

Let $ M$ be number of such sequences consisting of $ k$ steps, resulting in the state where lamps $ 1$ through $ n$ are all on, and lamps $ n + 1$ through $ 2n$ are all off, but where none of the lamps $ n + 1$ through $ 2n$ is ever switched on.

Determine $ \frac {N}{M}$.

:o

pythag: Trivial problems are trivial.

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Plagiarism: 2008 IMO #5

by Yongyi781, Apr 12, 2009, 8:09 PM

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eh, i was just waiting for pythag to post his solution before quoting the source.

by gauss1181, Apr 12, 2009, 9:05 PM

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pythag already said that 2008 imo was trivial except for number 6

by AIME15, Apr 15, 2009, 2:04 AM

Click here if you did not make white MOP (Fake copy of pythag's blog)

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