Given:

There are four gaps in between the girls where the boys can sit.

Let the number of boys in these gaps be

2a + 1,2b + 1,2c + 1,2d + 1

Then

2a + 1 + 2b + 1 + 2c + 1 + 2d +1 = 10

\(\Rightarrow a + b + c + d = 3 \)

The number of solution of above equation = coefficient of \( x^{3} in (1 – x)^{4} = 6_{C_{3}} = 20 \)

Here both the boys and girls can sit in 20 * 10! * 5! ways

Therefore, total ways = 15!

The required probability = k = 4.

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