Answer these questions independently of each other
6 men and 3 women are to be seated for a dinner, In a row, such that no two women sit together. Find the number of ways in which this arrangements can be done.

Options

A. 121500
B. 151200
C. 112500
D. `6!xx120`

Answer

To solve the problem of seating 6 men and 3 women in a row such that no two women sit together, we can follow these steps:

### Step 1: Arrange the Men
First, we will arrange the 6 men. The number of ways to arrange 6 men in a row is given by the factorial of the number of men:

\[ \text{Ways to arrange men} = 6! = 720 \]

### Step 2: Identify the Gaps for Women
Once the men are seated, we need to identify the gaps where the women can sit. When 6 men are seated, they create gaps where the women can be placed. The arrangement of 6 men creates 7 gaps (one before each man, one after the last man):

- Gap 1: Before Man 1
- Gap 2: Between Man 1 and Man 2
- Gap 3: Between Man 2 and Man 3
- Gap 4: Between Man 3 and Man 4
- Gap 5: Between Man 4 and Man 5
- Gap 6: Between Man 5 and Man 6
- Gap 7: After Man 6

### Step 3: Choose Gaps for Women
We need to choose 3 out of these 7 gaps to place the women, ensuring that no two women sit together. The number of ways to choose 3 gaps from 7 is given by the combination formula:

\[ \text{Ways to choose gaps} = \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

### Step 4: Arrange the Women
After choosing the gaps, we can arrange the 3 women in those chosen gaps. The number of ways to arrange 3 women is given by:

\[ \text{Ways to arrange women} = 3! = 6 \]

### Step 5: Calculate Total Arrangements
Now, we can find 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 calculated:

\[ \text{Total arrangements} = 720 \times 35 \times 6 \]

Calculating this step-by-step:

1. \(720 \times 35 = 25200\)
2. \(25200 \times 6 = 151200\)

Thus, the total number of ways to arrange 6 men and 3 women such that no two women sit together is:

\[ \boxed{151200} \]