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 arrange the 6 students. Since all students are distinct, the number of ways to arrange them 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 seated. The gaps are as follows:

- Before the first student
- Between the first and second students
- Between the second and third students
- Between the third and fourth students
- Between the fourth and fifth students
- Between the fifth and sixth students
- After the last student

This gives us a total of 7 gaps.

### Step 3: Choose Gaps for Teachers
We need to select 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:

\[ \text{Ways to choose gaps} = \binom{7}{4} = 35 \]

### Step 4: Arrange the Teachers
After selecting the gaps, we can arrange the 4 teachers in the chosen gaps. Since all teachers are distinct, the number of ways to arrange them is given by the factorial of the number of teachers.

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

### Step 5: Calculate the Total Arrangements
Finally, we multiply 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:

\[ \text{Total arrangements} = (6!) \times \binom{7}{4} \times (4!) = 720 \times 35 \times 24 \]

Calculating this gives:

\[ 720 \times 35 = 25200 \]
\[ 25200 \times 24 = 604800 \]

Thus, the total number of ways to arrange 6 students and 4 teachers in a row such that no two teachers are together is **604800**.

### Final Answer:
The total number of arrangements is **604800**.

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