The number of ways in which m men and n women can be seated in a row, so that no two women sit together is
(i) `(m!(m+1)!)/((m+n-1)!)`
(ii) `(m!(m+1)!)/((m-n+1)!)`
(iii) `(n!(m+1)!)/((m-n+1)!)`
(iv) `(m!(n+1)!)/((m+n-1)!)`

Options

A. `(|ul(m)|ul(m+1))/(|ul(m+n-1))`
B. `(|ul(m)|ul(m +1))/(|ul(m-n+1))`
C. `(|ul(n)|ul(m+1))/(|ul(m-n+1))`
D. `(|ul(m)|ul(n+1))/(|ul(m+n-1))`

Answer

To solve the problem of seating \( m \) men and \( n \) women in a row such that no two women sit together, we can follow these steps:

### Step 1: Arrange the Men
First, we arrange the \( m \) men. The number of ways to arrange \( m \) men in a row is given by \( m! \).

**Hint:** Remember that the factorial of a number \( n \) (denoted as \( n! \)) represents the product of all positive integers up to \( n \).

### Step 2: Identify the Positions for Women
Once the men are arranged, we need to identify the positions where the women can sit. When \( m \) men are seated, they create \( m + 1 \) potential gaps for the women to occupy (one before each man, one after the last man).

**Hint:** Visualize the arrangement of men to see how many gaps are created for the women.

### Step 3: Choose Positions for Women
We need to select \( n \) gaps from the \( m + 1 \) available gaps to seat the \( n \) women. The number of ways to choose \( n \) gaps from \( m + 1 \) is given by the combination formula \( \binom{m+1}{n} \).

**Hint:** The combination formula \( \binom{n}{r} \) is calculated as \( \frac{n!}{r!(n-r)!} \).

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

**Hint:** Again, use the factorial concept for arranging the women.

### Step 5: Combine the Results
The total number of arrangements is the product of the arrangements of men, the selection of gaps, and the arrangements of women. Thus, the total number of ways to arrange \( m \) men and \( n \) women such that no two women sit together is given by:

\[ \text{Total Ways} = m! \times \binom{m+1}{n} \times n! \]

Substituting the combination formula:

\[ \text{Total Ways} = m! \times \frac{(m+1)!}{n!(m+1-n)!} \times n! \]

This simplifies to:

\[ \text{Total Ways} = m! \times \frac{(m+1)!}{(m+1-n)!} \]

### Final Answer
Thus, the total number of ways in which \( m \) men and \( n \) women can be seated in a row such that no two women sit together is:

\[ \frac{m!(m+1)!}{(m-n+1)!} \]

This corresponds to option (ii).