Clara is afraid to forget the secret combination of numbers that opens the safe. Therefore, she decided to write this combination in a notebook in encrypted form. For encryption, the numbers were written out in a table, after which Clara rearranged the columns of this table at random several times, remembering the permutation method. Then she rearranged the columns again in the same way and wrote down the final result in a notebook. For what minimum number = in brute force combinations can you open the safe?

4 5 6 2 2 0 2 9 0 1 9 9

Like a solution to the forehead

p = 6! = 720 combinations.

But the teacher assures that there is a solution in which the number of combinations will be noticeably less. Through numerous attempts to surrender the task, find out which way to think, and what I am doing wrong, in response, I achieved only that the value is greater than 256 and less than 360. Any ideas?

  • I do not know where, but I got 6*5*5 + 6*5*4 + 6*5*3 + 6*5*2 + 6*5*1 = 360 :-D although probably the first combination will be 6.5.4 ... that is, the result is 6*5*4 + 6*5*3 + 6*5*2 + 6*5*1 = 285 - Alexey Shimansky
  • as I tried to explain, I need to calculate the squares of all permutations. The squares are among the even permutations, it means no more than 360. From this number it is necessary to subtract the number of those even permutations that are not squares. But I did not understand anything - lion295

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