In how many ways may $5$ people sit in a row of five chairs if two of them refuse to sit next to each other?

Question

I think I have a hard problem. The problem is:

In how many ways may $5$ people sit in a row of five chairs if two of them refuse to sit next to each other?

The hard part is the possibilities of them not sitting next to each other.

Answer 1

Figure out the number of ways you can place those two people, and then add the other three.

For the first part, there are $ 6 $ pairs of seats not next to each other and $2 $ ways to place those two people in those two seats, so that's $ 12 $ ways to seat those two people

Then placing the other three can be done in $3!=6$ ways

Total: $12 \cdot 6=72$ ways to seat those 5 people

Answer 2

Note that

A) The ways for 5 people to sit in a row of five is $5!$.

B) The ways two of them are next to each other is $4\cdot2$ and for each case you have 3! ways for the Others, thus we have $8\cdot 3!$ forbidden configurations.

Therefore the number of ways in which 5 people can sit in a row of five chairs if two of them refuse to sit next to each other are given by $A-B$ that is

$$5!-8\cdot 3!=120-48=72$$

Answer 3

Comment: Not a different answer, but maybe a (slightly) different way to think about the problem:

To find the number of arrangements in which the feuding two sit together, consider arrangements of four objects: this pair and the other three people, then permute the two in the pair. $2(4!) = 48.$

The number of arrangements without restriction is $5! =120.$

The answer is the difference of these two numbers, which is @gemusi's Answer. (+1)

I think this method is 'best' because it easily generalizes to $n \ge 3$ people.