Six boys and six girls sit along a line alternately in x ways, and along a circle (again alternatively in y ways), then

Options

A. `x=y`
B. `y=12x`
C. `x=10y`
D. `x=12y`

Answer

To solve the problem of arranging six boys and six girls alternately in a line and in a circle, we will break it down step by step.

### Step 1: Arranging in a Line
1. **Arranging Alternately**: We can start with either a boy or a girl.
- If we start with a boy, the arrangement will be: Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl.
- If we start with a girl, the arrangement will be: Girl, Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl, Boy.

2. **Calculating Arrangements**:
- For the arrangement starting with a boy:
- The boys can be arranged in \(6!\) ways.
- The girls can also be arranged in \(6!\) ways.
- Therefore, the total arrangements for this case is:
\[ 6! \times 6! \]
- For the arrangement starting with a girl:
- The arrangement is the same, so it will also be:
\[ 6! \times 6! \]
- Thus, the total arrangements in a line (denoted as \(x\)) is:
\[ x = 6! \times 6! + 6! \times 6! = 2 \times (6!)^2 \]

### Step 2: Arranging in a Circle
1. **Arranging Alternately in a Circle**: In a circular arrangement, we can fix one person to eliminate the effect of rotations.
- Let's fix one boy. The arrangement will then look like this: Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl.
- This means we will have 5 remaining boys and 6 girls to arrange.

2. **Calculating Arrangements**:
- The remaining 5 boys can be arranged in \(5!\) ways (since one boy is fixed).
- The girls can still be arranged in \(6!\) ways.
- Therefore, the total arrangements in a circle (denoted as \(y\)) is:
\[ y = 5! \times 6! \]

### Step 3: Finding the Relation Between \(x\) and \(y\)
1. **Setting Up the Equation**:
- We have:
\[ x = 2 \times (6!)^2 \]
- and
\[ y = 5! \times 6! \]

2. **Dividing \(x\) by \(y\)**:
\[ \frac{x}{y} = \frac{2 \times (6!)^2}{5! \times 6!} \]
- Simplifying this gives:
\[ \frac{x}{y} = \frac{2 \times 6!}{5!} \]
- Since \(6! = 6 \times 5!\), we can substitute:
\[ \frac{x}{y} = \frac{2 \times 6 \times 5!}{5!} = 2 \times 6 = 12 \]

3. **Final Relation**:
- Thus, we find that:
\[ x = 12y \]

### Conclusion
The relation between \(x\) and \(y\) is:
\[ x = 12y \]