m women and n men are too be seated in a row so that no two men sit together. If `mgtn` then show that the number of wys in which they can be seated is `(m!(m+1)!)/((m-n+1)!)`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Arrangement**:
We have `m` women and `n` men. Since no two men can sit together, we need to arrange the women first. By arranging the `m` women, we create spaces for the men.

2. **Arranging the Women**:
The `m` women can be arranged in `m!` ways.

\[ \text{Ways to arrange women} = m! \]

3. **Identifying Spaces for Men**:
When the `m` women are seated, they create `m + 1` potential spaces for the men. This includes:
- One space before the first woman
- One space between each pair of women
- One space after the last woman

Therefore, the total number of spaces available for the men is `m + 1`.

4. **Choosing Spaces for Men**:
We need to choose `n` spaces out of the `m + 1` available spaces for the `n` men. The number of ways to choose `n` spaces from `m + 1` is given by the combination formula:

\[ \text{Ways to choose spaces} = \binom{m+1}{n} = \frac{(m+1)!}{n!(m+1-n)!} \]

5. **Arranging the Men**:
Once we have chosen the spaces, the `n` men can be arranged in those chosen spaces in `n!` ways.

\[ \text{Ways to arrange men} = n! \]

6. **Combining the Arrangements**:
Now, we combine all the arrangements together. The total number of ways to arrange the `m` women and `n` men such that no two men sit together is:

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

Substituting the values we calculated:

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

The `n!` in the numerator and denominator cancels out:

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

Simplifying this gives:

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

### Final Result:
Thus, the number of ways to arrange `m` women and `n` men such that no two men sit together is:

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