Question

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

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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.

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.
Hence, the total no. of ways we can sit 5 persons with 2 persons always together is $6 \times 2 = 12$ ways ...(2)
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.