The number of ways in which we can arrange n ladies and n gentlemen at a round table so that 2 ladies and 2 gentlemen may not sit next to one another is:
\[
  a.{\text{ }}\left( {n - 1} \right)!\left( {n - 2} \right)! \\
  b.{\text{ }}\left( {{\text{n!}}} \right)\left( {n - 1} \right)! \\
  c.{\text{ }}\left( {n + 1} \right)!\left( {n!} \right) \\
  d.{\text{ None of these}} \\
\]

Question

The number of ways in which we can arrange n ladies and n gentlemen at a round table so that 2 ladies and 2 gentlemen may not sit next to one another is:
\[
  a.{\text{ }}\left( {n - 1} \right)!\left( {n - 2} \right)! \\
  b.{\text{ }}\left( {{\text{n!}}} \right)\left( {n - 1} \right)! \\
  c.{\text{ }}\left( {n + 1} \right)!\left( {n!} \right) \\
  d.{\text{ None of these}} \\
\]

Vedantu Content Team · Accepted Answer

Hint: - Number of ways to sit n persons on a round table is $ = \left( {n - 1} \right)!$

First we have to arrange n gentlemen around the round table so, number of ways to do so
$ = \left( {n - 1} \right)!$
Now, when these men are arranged and seated than there are n spaces between each man where we will arrange and seated n ladies so that two ladies and two gentlemen may not sit each other$ = {}^n{C_n}\left( {n!} \right)\left( {n - n} \right)! = {}^n{C_n}\left( {n!} \right)\left( {0!} \right)$
As we know the value of ${}^n{C_n} = n,{\text{ }}0! = 1$
$ \Rightarrow {}^n{C_n}\left( {n!} \right)\left( {0!} \right) = 1 \times n! \times 1 = n!$
Hence, total number of ways of sitting so that two ladies and two gentlemen may not sit each other
$ = \left( {n - 1} \right)!\left( {n!} \right)$
Hence, option (b) is correct.

Note: - Whenever we face such types of problems first calculate the number of ways to sit $n$gentlemen on a round table, then calculate the number of ways to sit $n$ ladies between them, then multiply these two values we will get the required answer.

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