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Difference between revisions of "2008 IMO Problems/Problem 5"
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What we want to show now is that each element of <math>\cal{M}</math> is an image of exactly <math>2^{k-n}</math> ekements from <math>cal{N}</math>, which would imply <math>N = 2^{k-n}M</math> and solve the problem.
 
What we want to show now is that each element of <math>\cal{M}</math> is an image of exactly <math>2^{k-n}</math> ekements from <math>cal{N}</math>, which would imply <math>N = 2^{k-n}M</math> and solve the problem.
  
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Consider an element <math>y</math> of <math>\cal{M}</math> and let <math>l_i</math> be the number of apearances of the number <math>i</math> in <math>y</math> for <math>i=1,2,\ldots n</math>.
+
Consider an element <math>y</math> of <math>\cal{M}</math> let <math>l_i</math> be the number of apearances of the number <math>i</math> in <math>y</math> for <math>i=1,2,\ldots n</math>. Now consider the set of pre-images of <math>y</math>, that is
 +
<cmath>X_y = \{ x \in \cal{N} | \math{f(x) = y }\}.</cmath>
 +
It is easy to see that each element <math>x\in X_y</math> is derived from <math>y</math> by ''flipping'' an ''even'' number of its <math>1</math>-s, <math>2</math>-s, and so on, where flipping means changing the number <math>k\in A</math> to <math>k+n\in B</math>. Since each such flipping results in a unique <math>x</math>, all we want to count is count the flippings. We can flip <math>0, 2, 4,\ldots </math> of the <math>1</math>-s and choose which ones, so that results in
 +
<cmath>\binom{l_1}{0} + \binom{l_1}{2}+\binom{l_1}{4}+\cdots+ = 2^{l_1-1}</cmath>
 +
flippings. Combine each of them with the <math>2^{l_2-1}</math>, <math>2^{l_3-1}</math>, etc. ways of flipping the <math>2</math>-s, <math>3</math>-s etc. respectively to get the total number of flippings
 +
<cmath>2^{(l_1-1)+(l_2-1)+\cdots+(l_n-1)} = 2^{l_1+l_2+\cdots+l_k-n} = 2^{k-n}.</cmath>
 +
This shows that <math>|X_y| = 2^{k-n}</math> and the proof is complete.

Revision as of 05:10, 4 September 2008

Problem 5

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 $\frac {N}{M}$.

Solution

For convenience, let $A$ denote the set $(1,2,\ldots n)$ and $B$ the set $(n+1,n+2,\ldots,2n)$.

We can describe each sequences of switching the lamps as a $k$-dimensional vector $(a_1, a_2, \ldots, a_k)$, where $a_i \in A \cup B$ signifies which lamp was switched on the $i$-th move for $i=1,2,\ldots k$.

Let $\cal{N}$ consist of those sequences that contain each of the numbers in $A$ an odd number of times and each of the numbers in $B$ an even number of times. Similarly, let $\cal{M}$ denote the set of those sequences that contain no numbers from $B$ and each of the numbers in $A$ an odd number of times. By definition, $M=|\cal{M}$ and $N=\cal{N}$.

Define the mapping $f:\cal{N}->\cal{M}$ as \[f(a_1, a_2, \ldots, a_k) = (b_1,b_2,\ldots b_k) : b_i = \left\{ \begin{array}{l} a_i \textrm{ if } a_i \in A \\ a_i-n \textrm{ if } a_i \in B \end{array} \right .\]

What we want to show now is that each element of $\cal{M}$ is an image of exactly $2^{k-n}$ ekements from $cal{N}$, which would imply $N = 2^{k-n}M$ and solve the problem.

Consider an element $y$ of $\cal{M}$ let $l_i$ be the number of apearances of the number $i$ in $y$ for $i=1,2,\ldots n$. Now consider the set of pre-images of $y$, that is 

\[X_y = \{ x \in \cal{N} | \math{f(x) = y }\}.\] (Error compiling LaTeX. Unknown error_msg)

It is easy to see that each element $x\in X_y$ is derived from $y$ by flipping an even number of its $1$-s, $2$-s, and so on, where flipping means changing the number $k\in A$ to $k+n\in B$. Since each such flipping results in a unique $x$, all we want to count is count the flippings. We can flip $0, 2, 4,\ldots$ of the $1$-s and choose which ones, so that results in \[\binom{l_1}{0} + \binom{l_1}{2}+\binom{l_1}{4}+\cdots+ = 2^{l_1-1}\] flippings. Combine each of them with the $2^{l_2-1}$, $2^{l_3-1}$, etc. ways of flipping the $2$-s, $3$-s etc. respectively to get the total number of flippings \[2^{(l_1-1)+(l_2-1)+\cdots+(l_n-1)} = 2^{l_1+l_2+\cdots+l_k-n} = 2^{k-n}.\] This shows that $|X_y| = 2^{k-n}$ and the proof is complete.

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