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.

Options

A. ` 6! Xx 5! `
B. ` 30 `
C. ` 5! Xx 4!`
D. ` 7! Xx 5! `

Answer

To solve the problem of arranging 6 men and 5 women at a round table such that no two women sit together, we can follow these steps:

### Step 1: Arrange the Men
First, we need to arrange the 6 men at the round table. In circular permutations, the number of ways to arrange \( n \) objects is given by \( (n-1)! \). Therefore, the number of ways to arrange 6 men in a circular manner is:

\[ (6 - 1)! = 5! = 120 \]

### Step 2: Identify the Gaps for Women
Once the men are seated, they create gaps where the women can sit. For 6 men, there will be 6 gaps (one between each pair of men and one after the last man).

### Step 3: Place the Women
Since we want to ensure that no two women sit together, we can only place one woman in each gap. We have 5 women to place in these 6 gaps. The number of ways to choose 5 gaps from the available 6 is given by \( \binom{6}{5} \), which is equal to 6.

### Step 4: Arrange the Women
After choosing the 5 gaps, we can arrange the 5 women in those gaps. The number of ways to arrange 5 women is given by \( 5! \).

### Step 5: Calculate the Total Arrangements
Now, we can calculate the total number of arrangements by multiplying the number of ways to arrange the men, the number of ways to choose the gaps, and the number of ways to arrange the women:

\[ \text{Total arrangements} = \text{Ways to arrange men} \times \text{Ways to choose gaps} \times \text{Ways to arrange women} \]

Substituting the values we found:

\[ \text{Total arrangements} = 5! \times \binom{6}{5} \times 5! = 120 \times 6 \times 120 \]

Calculating this gives:

\[ \text{Total arrangements} = 120 \times 6 \times 120 = 86400 \]

### Final Answer
Thus, the number of ways in which 6 men and 5 women can dine at a round table such that no two women sit together is:

\[ 86400 \]

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