Question

The number of ways in which 5 ladies and 7 gentlemen can be seated in a round table so that no two ladies sit together, is:
1) \[\dfrac{7}{2}{(720)^2}\]
2) \[7{(360)^2}\]
3) \[7{(720)^2}\]
4) \[720\]
5) \[360\]

Answer 1

Hint: We can solve the given simple problem by using the concept of circular permutation
And we know that the formula for circular permutation and is given by \[\dfrac{{n!}}{n} = (n - 1)!\].since there are 5 ladies and 7 gentlemen so first, we allocate gentlemen and then we can allocate ladies in the available places so that no two of them sit together.

Complete step-by-step solution:
Since there are 5 ladies and 7 gentlemen to be seated at a round table
7 gentlemen can be seated in \[(n - 1)!\] \[ = (7 - 1)!\] as per the formula of circular permutation.
Therefore, We can arrange 7 gentlemen in 6! Ways
Whenever we are solving the problems related to arrangement, we should use the concept of permutation.
Now we have to arrange 5 ladies in available 7 places so that no two of them sit together is \[7{p_5}\] ways
Now using the formula of permutation is given by \[n{p_r} = \dfrac{{n!}}{{(n - r)!}}\]
Replace n by 7 and r by 5 we get
\[7{p_5} = \dfrac{{7!}}{{(7 - 5)!}} = \dfrac{{7!}}{{2!}}\]
Whenever we have keywords ‘and’ in the problem we have to multiply. That is we have to arrange both gentlemen and ladies
Therefore,
The number of ways in which ladies and gentlemen can be seated in a round table so that no two ladies sit together, is = number ways of arranging ladies \[ \times \] number ways of arranging gentlemen
\[ \Rightarrow {\text{6! }}\times\dfrac{{{\text{7!}}}}{{{\text{2!}}}}\]
We can rewrite it as
\[ \Rightarrow 6! \times \dfrac{{7 \times 6!}}{{2!}}\]
On simplification we get
\[ \Rightarrow 720 \times \dfrac{{7 \times 720}}{2}\]
\[ \Rightarrow \dfrac{7}{2}{(720)^2}\]
Therefore, the correct answer is option 1) \[\dfrac{7}{2}{(720)^2}\]

Note: Permutation in a circle is called circular permutation. If we consider a round table and 3 persons then the number of different sitting arrangements that we can have around the round table is an example of circular permutation. Now for n elements, circular permutation \[\dfrac{{n!}}{n} = (n - 1)!\]
If the clockwise and counterclockwise orders can be distinguished then the total number of circular permutations of n elements taken all together = (n-1)!
If the clockwise and counter clockwise orders cannot be distinguished then total number of circular permutations of n elements taken all together = (n-1)! / 2