Question

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by
A. \[6! \times 5!\]
B. \[30\]
C. \[5! \times 4!\]
D. \[7! \times 5!\]

Answer 1

Hint: In this question, we need to find the total number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together. For this, we need to use the concept of factorial and mathematical identities for the factorial of a number and the permutation.

Formula used: We will use the following mathematical rule for the factorial of a number.
\[n! = n \times \left( {n - 1} \right) \times .... \times 1\]
Here, \[n\] is a positive integer.
Also, the permutation rule is given by
\[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Here, \[n\] and \[r\] are positive integers.

Complete step-by-step solution:
We know that there are 6 men and 5 women.
Here, we can place 5 women in 6 empty places between them so that no two women will be together.
So, this can be done in the following way.
Thus, we get
\[{}^6{P_5} = \dfrac{{6!}}{{\left( {6 - 5} \right)!}}\]
By simplifying, we get
\[{}^6{P_5} = \dfrac{{6!}}{{\left( 1 \right)!}}\]
\[{}^6{P_5} = 6!\]
\[{}^6{P_5} = 6 \times 5 \times 4 \times 3 \times 2 \times 1\]
\[{}^6{P_5} = 720\]
Since the activities are interdependent, the number of ways in which 6 men and 5 women can dine at a round table if no two women sit together is limited.
Thus, we get
\[6! \times 5! = 720 \times 5 \times 4 \times 3 \times 2 \times 1\]
\[6! \times 5! = 720 \times 120\]
By simplifying, we get
\[6! \times 5! = 86400\]
Hence, there are 86400 ways in which 6 men and 5 women can dine at a round table if no two women are to sit together.
That is \[6! \times 5!\] ways.

Therefore, the correct option is (A).

Note: Many students make mistakes in the permutation formula. They may confuse about the combination and permutation formulae. Due to the very small difference between the two formulae, we get a big difference in the desired result.