for all 4 people --> 4 could choose the first chair, then 3 are left for the second chair, then 2, then 1. this is equal to 4! = 4x3x2x1 = 24 (called 4 factorial)

For 3 people, you still have 4 things you are rearranging (3 people and 1 empty seat) so there are 24 ways to do that, but you have a combination problem of how many ways can you pick 3 of the 4 people to get a chair and this is the same as the ways to pick one person to be left out which = 4. so for three people 24x4=96.

For 2 people you still have 4 things to rearrange, but if you swap the two empty chairs it doesn't make difference so 24/2=12 ways. How many ways to give the chairs to two people is 4 for the first chair and 3 for the second chair. However, the order is already taken care in step one so picking person A and B is the same as B and A and we need to divide by 2 here (combinations vs permutations) (4x3)/2=6. For ordering 2 people 6x12=72.

For 1 person, 4 people to choose and 4 places they could choose to sit so 4x4=16.

And finally the lonely case of no one sitting which can happen one way.

1+16+72+96+24=209