March’s Problem of the Month was a tough one on permutations. Only six people solved it! (Site: http://www.fen.bilkent.edu.tr/~cvmath/Problem/problem.htm)
The question goes as follows:
In each step one can choose two indices and transform the 100 tuple
into the 100 tuple
if
is an even number. We say that a permutation
of
is good if starting from
one can obtain it after finite number of steps. Find the total number of distinct good permutations of
.
The official solution is beautiful and uses induction.
Personally, I used a more brute force technique to get the same answer using equivalence class theory which I learnt in my first year of undergraduate math! It is not so bad in this question, since n is only 100, but for higher values of n the approach in the official solution would be better.
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Haha, I would have also use the equivalence class theory to solve this question if I encountered it. But yeah, it would be ridiculously tedious to solve it if n is higher…
I’ll have a closer look to their official solution, it seems interesting.
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Yeah, their official solution is awesome! It didn’t cross my mind to use the cyclic permutation, so I solved it using a more tedious way.
Thanks for commenting!
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Yeah, that approach was very nice indeed
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