(2n) and below (2n-2) ... 42 (2n-1) (2n-3) ... 31all constitute the reverse order, with 2n-1; (2n-2) and the following (2n-4) 42 (2n-3)...3 1 all constitute an inverse order, with 2n-3; 4 and the following 23 1 constitute the reverse order, and there are three; 2 and the following 1 form the reverse order, with 1.
(2n- 1) and the following (2n-3)…3 1 all constitute the reverse order, where n-1; (2n-3) and below (2n-5)...3 1 all constitute the reverse order, with n-2; …, 5 and the following 3 1 constitute the reverse order, and there are two; 3 and the following 1 form the reverse order, with 1. So the inverse number is [(2n-1)+(2n-3)+…+3+1]+[(n-1)+…+2+1] = n (3n-/kloc-0
In one arrangement,
If the front and back positions of a pair of numbers are in reverse order, that is, the front number is greater than the back number, it is called reverse order. The total number of inverses in an arrangement is called the number of inverses in this arrangement.
For n different elements, it is stipulated that there is a standard order between elements (for example, it can be stipulated that n different natural numbers have a standard order from small to large), so in any arrangement of these n elements, when the actual order of two elements is different from the standard order, it is said that there is 1 reverse order. The total number of all inverses in an arrangement is called the number of inverses in this arrangement.