Let's say.....

2 women and 3 men are to sit in a row. Find in how many ways they
can be arranged so that the two women do not sit next to each other.

You need to learn to use factorials

@davitchen You are welcome lol

@bunyonb Im also doing exam in the next week I'll have to repeat. So it was helpful that post ahahaha

@davitchen Studying. I need to understand it and other similar problems for exam prep.

@bunyonb Well thanks, have you been studying, or is it just curiosity?

@davitchen Wow that's great man.

It also can be done with permutations.
Personally i think it's easier to make 5P5  (4P4 x 2) = 120  48 = 72 also

@GirlNextDoor Who?

This guy is a computer scientist and it he has a amazing mind

@Concrete_Angel lol np

@bunyonb oh I feel quite embarrassed haha as I have just realised that I misread the question I read it as 4 men not 3 so my working is based on there being 6 people not 5.
5!  (4! × 2) = 120  48 = 72
So there are actually 72 combinations, a big difference to my first answer! Sorry about that haha

@Concrete_Angel awww ok

@bunyonb I wouldn't know if u were missing out rules without knowing your working towards the problem and also I'm not overly familiar with the subject myself haha

@Concrete_Angel I can never seem to get the hang of combinatorics. Are there some rules that im missing out?

There are 720 ways in which 6 people can be seated in a row. However 240 of those ways would have the 2 women next to eachother so subtract 240 from 720 and the answer is 480. There are 480 ways in which they can be seated without the women being next to eachother.
6!  (5! × 2) = 720240 = 480.

@hannah Nice

@bunyonb 3x2 = 6 2 = 4. 4 ways.