# Concrete Mathematics Summation Question

I'm sorry if this question is too novice, but I am just beginning discrete math. I've been working through the book Concrete Mathematics (Graham,Knuth,Patashnik) and I reached a double summation that has me very confused. I'v been trying to work it out, and I think I have a solution but I'm not sure if it is the correct way to solve it.

The question comes from chapter 2 section 4 and it goes as follows: $$S = \displaystyle\sum\limits_{1 \le j < k \le n}^{}{(a_k - a_j)(b_k - b_j)}$$

The authors then go on to say " We have symmetry when j and k are interchanged:" and write the new sum: $$S = \displaystyle\sum\limits_{1 \le k < j \le n}^{}{(a_j - a_k)(b_j - b_k)} = \displaystyle\sum\limits_{1 \le k < j \le n}^{}{(a_k - a_j)(b_k - b_j)}$$ I understand how they can get to the 2nd sum, because all you're doing is changing index names. But how do the authors get from the 2nd sum to the 3rd sum? Or how do they use their previously mentioned "Rocky Road" formula to achieve this result?

Thanks,

EDIT: Sorry for not making the Rocky Road formula clear. The Rocky Road Formula is as follows:

$$\displaystyle\sum\limits_{j \in J}^{}{\displaystyle\sum\limits_{k \in K(j)}^{}{a_j,_k}} = \displaystyle\sum\limits_{k \in K'}^{}{\displaystyle\sum\limits_{j \in J'(k)}^{}{a_j,_k}}$$

All we're doing in moving from the second sum to the third sum is essentially multiplying each bracket by $-1$, but since there are two brackets, the two factors of $-1$ cancel out so the sum is unchanged.
In this book author say with Iversonian notation if $$P(j,k)=[j \in J][k \in K(j)]= [k \in K^{'}][j \in J^{'}(k)]$$ then we can use any of this statements for index of sum. for example $$P(j,k)= [1 \leqslant‎ j \leqslant‎ k \leqslant‎ n]=[1 \leqslant‎ j \leqslant‎ n][j \leqslant‎ k \leqslant‎ n] =[1 \leqslant‎ k \leqslant‎ n][1 \leqslant‎ j \leqslant‎ k]$$ then the summation with single index $$1 \leqslant‎ j \leqslant‎ k \leqslant‎ n$$ is equal to double summations with two other perdicates.