A lock with three buttons opens if the buttons are pressed in a predetermined order, one after the other. What is the minimum number of button presses needed to ensure that the lock will open? (The correct sequence of three button presses is not affected by any preceding button presses.)





Translating the problem statement into English as requested, while preserving the original formatting and line breaks.
There are $3!=6$ different orders in which the three buttons can be pressed to unlock the lock, so a sequence of button presses is needed that contains all six possible orders. We will show that nine button presses are sufficient, but fewer than that are not. If we number the individual buttons with the numbers 1, 2, 3, then the following sequence is clearly suitable: 123121321.

Suppose there is also an eight-element sequence made from the numbers 1, 2, 3 that is suitable for unlocking the lock. Such a sequence allows exactly six attempts - thus, a shorter sequence is clearly not possible - so the triples starting at different positions are all different, and each is a permutation of the elements $\{1,2,3\}$. From this, it also follows that adjacent or second-adjacent elements cannot be equal.

Consider the element $h$ in the third position. Since there are two permutations starting with $h$, $h$ appears again in the sequence. This second occurrence cannot be in the first five or the last two positions, otherwise there would be adjacent equal elements, or the other permutation starting with $h$ would not "fit". Thus, $h$ appears for the second time in the sixth position.

The two permutations starting with $h$ can only be different if the fifth and seventh positions contain the same number. This, however, is not possible, as it would result in a repeated element in the triple starting at the fifth position. Indeed, there is no eight-element or shorter sequence that would definitely unlock the lock.

![](https://cdn.mathpix.com/cropped/2024_05_02_5119b24c647c295c1f22g-1.jpg?height=174&width=759&top_left_y=794&top_left_x=672)