4 baseball players each stand at different corners of a baseball diamo

I don't know why there is such a solution in OE. For me it's circular permutation for 4 which is (4-1)!=3!=6.

The number of circular permutations of n different objects is (n-1)!.

Each circular permutation corresponds to n linear permutations depending on where we start. Since there are exactly n! linear permutations, there are exactly n!/n permutations. Hence, the number of circular permutations is the same as (n-1)!.

When things are arranged in places along a line with first and last place, they form a linear permutation. When things are arranged in places along a closed curve or a circle, in which any place may be regarded as the first or last place, they form a circular permutation.

The permutation in a row or along a line has a beginning and an end, but there is nothing like beginning or end or first and last in a circular permutation. In circular permutations, we consider one of the objects as fixed and the remaining objects are arranged as in linear permutation.

Thus, the number of permutations of 4 objects in a row = 4!, where as the number of circular permutations of 4 objects is (4-1)! = 3!.