Answer these questions independently of each other
5 men and 4 women are to be seated for a dinner, In a row, such that no two men sit together. Find the number of ways in which this arrangements can be done.

Options

A. 1440
B. 2880
C. 1720
D. 720

Answer

To solve the problem of seating 5 men and 4 women in a row such that no two men sit together, we can follow these steps:

### Step 1: Arrange the Women
First, we will arrange the 4 women. The number of ways to arrange 4 women is given by the factorial of the number of women:

\[ \text{Ways to arrange women} = 4! = 24 \]

### Step 2: Identify the Gaps for Men
Once the women are arranged, they will create gaps where the men can be seated. For 4 women, the arrangement will look like this:

W1 _ W2 _ W3 _ W4

Here, "W" represents a woman, and "_" represents a gap. There are 5 gaps available (one before each woman and one after the last woman):

1. Before W1
2. Between W1 and W2
3. Between W2 and W3
4. Between W3 and W4
5. After W4

### Step 3: Place the Men in the Gaps
Since we have 5 men and 5 gaps, we can place one man in each gap. The number of ways to arrange 5 men in these 5 gaps is given by:

\[ \text{Ways to arrange men} = 5! = 120 \]

### Step 4: Calculate the Total Arrangements
Now, we multiply the number of arrangements of women by the number of arrangements of men to find the total arrangements:

\[ \text{Total arrangements} = \text{Ways to arrange women} \times \text{Ways to arrange men} = 4! \times 5! = 24 \times 120 = 2880 \]

### Final Answer
Thus, the total number of ways to arrange 5 men and 4 women in a row such that no two men sit together is:

\[ \boxed{2880} \]

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