Difference between revisions of "2001 IMO Shortlist Problems/C6"
(New page: == Problem == For a positive integer <math>n</math> define a sequence of zeros and ones to be [i]balanced[/i] if it contains <math>n</math> zeros and <math>n</math> ones. Two balanced sequ...) |
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== Problem == | == Problem == | ||
− | For a positive integer <math>n</math> define a sequence of zeros and ones to be | + | For a positive integer <math>n</math> define a sequence of zeros and ones to be <i>balanced</i> if it contains <math>n</math> zeros and <math>n</math> ones. Two balanced sequences <math>a</math> and <math>b</math> are <i>neighbors</i> if you can move one of the <math>2n</math> symbols of <math>a</math> to another position to form <math>b</math>. For instance, when <math>n = 4</math>, the balanced sequences <math>01101001</math> and <math>00110101</math> are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set <math>S</math> of at most <math>\frac {1}{n + 1} \binom{2n}{n}</math> balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in <math>S</math>. |
== Solution == | == Solution == |
Latest revision as of 17:34, 20 August 2008
Problem
For a positive integer define a sequence of zeros and ones to be balanced if it contains zeros and ones. Two balanced sequences and are neighbors if you can move one of the symbols of to another position to form . For instance, when , the balanced sequences and are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set of at most balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in .
Solution
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