Let m denotes the number of ways in which 5 boys and 5 girls can be arranged in a line alternately and n denotes the number of ways in which 5 boys and 5 girls an be arranged in a circle so that no two boys are together . If m= kn then the value of k is :

Options

A. 2
B. 5
C. 6
D. 10

Answer

To solve the problem, we need to calculate the values of \( m \) and \( n \) and then find the value of \( k \) such that \( m = k \cdot n \).

### Step 1: Calculate \( m \)
\( m \) is the number of ways to arrange 5 boys and 5 girls in a line alternately.

1. **Arrange the boys**: There are 5 boys, and they can be arranged in \( 5! \) ways.
2. **Arrange the girls**: Similarly, the 5 girls can also be arranged in \( 5! \) ways.
3. **Alternating arrangement**: The arrangement can start with a boy or a girl. Therefore, we have two cases:
- Case 1: Boy first (BGBGBGBGBG)
- Case 2: Girl first (GBGBGBGBGB)

Thus, the total number of arrangements is:
\[ m = 5! \times 5! + 5! \times 5! = 2 \times 5! \times 5! \]

### Step 2: Calculate \( n \)
\( n \) is the number of ways to arrange 5 boys and 5 girls in a circle such that no two boys are together.

1. **Arrange the girls in a circle**: When arranging \( n \) objects in a circle, the number of arrangements is \( (n-1)! \). Hence, for 5 girls, the arrangements are:
\[ 4! \text{ (since we fix one girl to break the circular symmetry)} \]
2. **Place the boys**: Once the girls are arranged, there are 5 gaps between them where boys can be placed. Since no two boys can be together, we can place one boy in each gap. The boys can be arranged in these gaps in \( 5! \) ways.

Thus, the total number of arrangements is:
\[ n = 4! \times 5! \]

### Step 3: Relate \( m \) and \( n \)
We have:
\[ m = 2 \times 5! \times 5! \]
\[ n = 4! \times 5! \]

Now we need to find \( k \) such that:
\[ m = k \cdot n \]

Substituting the values of \( m \) and \( n \):
\[ 2 \times 5! \times 5! = k \cdot (4! \times 5!) \]

### Step 4: Simplify the equation
We can cancel \( 5! \) from both sides:
\[ 2 \times 5! = k \cdot 4! \]

Now, substituting \( 4! = 24 \):
\[ 2 \times 120 = k \cdot 24 \]
\[ 240 = k \cdot 24 \]

### Step 5: Solve for \( k \)
Now, divide both sides by 24:
\[ k = \frac{240}{24} = 10 \]

### Final Answer
The value of \( k \) is:
\[ \boxed{10} \]