In how many ways can 5 persons sit at a round table, if two persons do not sit together?

Hint: First, we will find the numbers of ways all persons can sit without any condition. Then, we will find the numbers of ways everyone can sit considering the persons sit together always and then subtract both to get the required answer.

Complete step-by-step answer:
Let us first get to know how to find the number of ways n persons can sit at a round table.
It is \[(n - 1)!\].
The reason for this is, if we have a straight line we have the answer as $n!$, but when it is transformed in round shape, then among those $n!$, n are the duplicate sitting arrangement of each other which means they are exactly n ways of showing one arrangement. Hence, it becomes \[\dfrac{{n!}}{n} = \dfrac{{n(n - 1)!}}{n} = (n - 1)!\].
Therefore, 5 persons can sit at a round table in $(5 - 1)! = 4! = 4 \times 3 \times 2 \times 1 = 24$ ……(1)
Now, if the two persons always sit together, we can consider them as one unit for once, so we now have to sit 4 people which can be done in $(4 - 1)! = 3! = 3 \times 2 \times 1 = 6$ ways.
But we must remember that the two persons who sat together can sit side by side in two ways, one in left or right of another person.
Hence, the total no. of ways we can sit 5 persons with 2 persons always together is $6 \times 2 = 12$ ways ...(2)
Now, subtracting (2) from (1), we will get:
Number of ways 5 persons sit at a round table, if two persons do not sit together = 24 – 12 = 12 ways
Hence, the answer is 12.

Note: The student might make the mistake of forgetting the fact that n persons can sit together in $(n - 1)!$ ways at a round table because n arrangements represent the same sitting arrangement. So, we have to exclude those.
If you directly approach the question, it will be really difficult to tackle the given condition. So, always try to break your question into parts if required, to make it easy.