In how many ways 6 students and 4 teachers be arranged in a row so that no two teachers are together ?

Options

A. 604800
B. 246800
C. 258600
D. 55500

Answer

To solve the problem of arranging 6 students and 4 teachers in a row such that no two teachers are together, we can follow these steps:

### Step 1: Arrange the Students
First, we need to arrange the 6 students. The number of ways to arrange 6 students is given by the factorial of the number of students:

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

### Step 2: Identify Positions for Teachers
Once the students are arranged, we need to find the positions available for the teachers. When the 6 students are arranged in a row, they create gaps where teachers can be placed.

For 6 students, the arrangement looks like this:

- S1 _ S2 _ S3 _ S4 _ S5 _ S6

Here, the underscores represent the gaps where teachers can be placed. There are a total of 7 gaps (one before each student and one after the last student).

### Step 3: Choose Positions for Teachers
We need to choose 4 out of these 7 gaps to place the teachers. The number of ways to choose 4 gaps from 7 is given by the combination formula:

\[ \text{Ways to choose gaps} = \binom{7}{4} \]

Calculating this:

\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

### Step 4: Arrange the Teachers
After choosing the gaps, we need to arrange the 4 teachers in the selected gaps. The number of ways to arrange 4 teachers is given by:

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

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

\[ \text{Total arrangements} = (6!) \times \binom{7}{4} \times (4!) \]

Substituting the values we calculated:

\[ \text{Total arrangements} = 720 \times 35 \times 24 \]

Calculating this step-by-step:

1. \(720 \times 35 = 25200\)
2. \(25200 \times 24 = 604800\)

Thus, the total number of ways to arrange 6 students and 4 teachers in a row such that no two teachers are together is:

\[ \text{Total arrangements} = 604800 \]

### Final Answer
The final answer is:

\[ \boxed{604800} \]