How many ways can you arrange 10 participants in a line so that two participants, Jack and Bob, aren't sitting next to each other?
I know that 10! is the no. of permutations for 10 participants, but I cannot understand how to do this with separating 2 individuals.
Consider first the number $N$ of configurations where Jack and Bob are sitting next each other. Let's consider for instance that Jack and Bob are conjoined twins.
Then you see that it is like arranging $9$ people the way you want, so you have $9 !$ possibilities. But now, Jack and Bob can be either the first before the second, either the second before the first. So for each possibility, you have two cases resulting of the position of Jack with respect to John.
The total number of configurations is then $N = 2 \times (9!)$.
The number of configurations where Jack and Bob are not sitting near each other is then the complement of $N$ to the total number of configurations, that is $$10 ! - 2 \times 9! = 8 \times 9 !$$