Two variants of the CAT paper are to be given to twelve students. In how many ways can the students be placed in two rows of six each so that there should be no identical variants side by side and that the students sitting one behind the other should have the same variant?

Options

A. `2 xx ""^(12)X_(6) xx (6 !)^2`
B. `6! xx 6!`
C. `7 ! xx 7 !`
D. None of these

Answer

To solve the problem of arranging 12 students in two rows of six each, with the conditions that no identical variants are side by side and that students sitting one behind the other have the same variant, we can follow these steps:

### Step 1: Understand the Arrangement
We have 12 students and we need to arrange them in two rows of 6 each. Each row will have students assigned to one of two variants (let's call them Variant A and Variant B).

### Step 2: Assign Variants
Since students sitting one behind the other must have the same variant, we can visualize the arrangement as follows:
- Row 1: A A A A A A (6 students with Variant A)
- Row 2: B B B B B B (6 students with Variant B)

### Step 3: Choose Students for Each Variant
We need to select 6 students out of the 12 to receive Variant A. The remaining 6 students will automatically receive Variant B. The number of ways to choose 6 students from 12 is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
In our case, \( n = 12 \) and \( r = 6 \):
\[ \binom{12}{6} = \frac{12!}{6!6!} \]

### Step 4: Arrange Students in Each Row
After selecting which students will receive Variant A and which will receive Variant B, we need to arrange the students in each row. Each row consists of 6 students, and the number of ways to arrange 6 students is given by \( 6! \). Therefore, for both rows, the total arrangements will be:
\[ 6! \times 6! \]

### Step 5: Calculate Total Arrangements
Now, we can combine all the parts together to find the total number of arrangements:
\[ \text{Total Arrangements} = \binom{12}{6} \times 6! \times 6! \]

### Step 6: Substitute and Calculate
Substituting the values we have:
\[ \binom{12}{6} = \frac{12!}{6!6!} = 924 \]
\[ 6! = 720 \]
Now, substituting these values into the total arrangements:
\[ \text{Total Arrangements} = 924 \times 720 \times 720 \]
Calculating this gives:
\[ \text{Total Arrangements} = 924 \times 720^2 \]
Calculating \( 720^2 = 518400 \):
\[ \text{Total Arrangements} = 924 \times 518400 = 479001600 \]

### Final Answer
Thus, the total number of ways to arrange the students under the given conditions is **479001600**.

---