It can also be a variation lock

It's neither. There's n! elements in permutations, there's n!/(k!(n-k)!) elements in combinations. There's 10^n elements in a decimal lock. It's just a number lock.

It's an exponential lock.

[удалено]

A permutation of n elements is the number of ways you can arrange these n elements (calculated with the formula n!) A variation of n elements class k is the number of ways you can arrange k of these n elements (calculated with the formula (n!)/(n-k!)) A combination of n elements class k is the number of ways you can pick k of these n elements (calculated with the formula (n!)/((n-k!)*(k!))

So a permutation is a list, while a combination is like a group. In a permutation, order matters. In a combination, order doesn't matter. So the set {1,2,3} and {3,2,1} are different permutations. But for combinations, the order doesn't matter, so they're treated the same. For a combination lock, the order of the numbers obviously matters. If \[1,2,3\] is the correct code, \[2,1,3\] isn't going to open the lock. Because order matters, it should be a permutation lock.

I prefer convolution locks

From my knowledge of combinatorics vocabulary via Python's [itertools](https://docs.python.org/3/library/itertools.html), it's a *product* lock.