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

Options

A. A. 604800
B. B. 24680
C. C. 25860
D. D. none of these

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 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 the Gaps for Teachers
Once the students are arranged, we need to identify the gaps where the teachers can be placed. When 6 students are arranged in a row, they create gaps where teachers can be placed. The gaps are as follows:

- Before the first student
- Between the students
- After the last student

For 6 students, there are a total of 7 gaps:

1. _ (before the first student)
2. S1 _ S2
3. S2 _ S3
4. S3 _ S4
5. S4 _ S5
6. S5 _ S6
7. _ (after the last student)

### Step 3: Choose Gaps for Teachers
We need to choose 4 gaps from these 7 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:

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

### Step 4: Arrange the Teachers
After choosing the gaps, we can arrange the 4 teachers in those 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 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:

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

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:

\[ \boxed{604800} \]