Question

The number of ways in which 6 boys and 4 girls can be arranged in a row such that between any two girls there must be exactly 2 boys is

• A. $6!5!$
• B. $(72)6!$
• C. $(144)5!$
• D. $4!7!$

Hint:

When a problem specifies a strict relative positioning of two groups (like "exactly X items between any two items of another group"), first determine the required pattern. If the pattern is fixed, the problem simplifies to arranging the members of each group within their designated slots.

Answer 1

The Correct Option is C

The condition is that between any two girls, there are exactly two boys.
Let G represent a girl and B represent a boy. With 4 girls, this condition fixes the arrangement pattern. The pattern must be: G B B G B B G B B G
This pattern uses exactly 4 girls and $2+2+2 = 6$ boys, which matches the numbers given in the problem.
No other arrangement is possible. For example, there cannot be boys at the ends, because then the end girl would not be "between two girls".
So, we have a fixed structure with 4 specific positions for girls and 6 specific positions for boys.
The 4 girls can be arranged among themselves in the 4 'G' positions in $4!$ ways.
The 6 boys can be arranged among themselves in the 6 'B' positions in $6!$ ways.
The total number of ways is the product of these two arrangements.
Total ways = $4! \times 6!$.
Now we need to match this with the given options.
$4! = 4 \times 3 \times 2 \times 1 = 24$.
$6! = 720$.
Total ways = $24 \times 720 = 17280$.
Let's evaluate option (C): $(144)5!$.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
$(144)5! = 144 \times 120 = 17280$.
Since $4! \times 6!$ is equal to $(144)5!$, option (C) is the correct answer.