[Algorithm Theory (Miscellaneous).] [ Processes and Operations ]

There are 6 locked suitcases and 6 keys to them. However, it is unknown which key fits which suitcase. What is the minimum number of attempts needed to definitely open all the suitcases? And how many attempts would be needed if there were not 6, but 10 suitcases and keys?

#
Try to determine in five attempts which of the 6 suitcases the first key fits.

## Solution

olnyaлись equalities in all rows and each

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![](https://cdn.mathpix.com/cropped/2024_05_06_e217990455cb7ab9a875g-06.jpg?height=74&width=50&top_left_y=620&top_left_x=2004)

![](https://cdn.mathpix.com/cropped/2024_05_06_e217990455cb7ab9a875g-06.jpg?height=40&width=35&top_left_y=677&top_left_x=2012)

## Solution

Standard incorrect solution: "Each of the six suitcases we try to open with each of the six keys, a total of $66=36$ attempts." We can find the correspondence between the keys and the suitcases in fewer attempts. We take the first key and try to open the suitcases one by one. If one of the suitcases opens, we set aside this suitcase with this key. If none of the first five suitcases open, then this key must correspond to the sixth suitcase. What happened? We used no more than five attempts; we have 5 keys and 5 suitcases left. We take one key again and open all the remaining suitcases in a row. To determine which suitcase this key corresponds to, we need four attempts. We take the next key and so on. In total, we need $5+4+3+2+1=15$ attempts. If there were 10 suitcases, the number of attempts would be $9+8+\ldots+2+1=45$.

## Answer

15 attempts; 45 attempts.

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