How many ways can 15 students be seated in a row such that the 2 talkative children never sit together?

The first thing to do is to work out how many ways there are without those 2 people minding. There are 5 possibilities for first place, 4 for second and so on, so the number of ways is 5x4x3x2x1 =...

It depends upon how many students are talkative. I'll give you two examples. 0 students talkative - 1 way 1 student talkative - 15 ways

There are (15 x 14) = 210 different ways to pull 2 from a bag of 15,but only 105 different pairs that you can end up with.

This is 6! {factorial} = 6 x 5 x 4 x 3 x 2 x 1 = 720. The way this works: take student A, who can choose any of the 6 desks, then for each of those 6 choices that student A makes, student B has 5...

(If they wouldn't want to sit together the answer would be 5! (120) but since there is a pair, we can count them as one person. So, my guess is that the answer is 4! (24)) The person who wrote this...