9 boys and 2 girls are to be placed next to each other in the school ground for morning assembly. What is the probability that there are exactly 4 boys between the 2 girls?

The total equals (9+2) = 11 individuals. Thus all possible arrangements is 11 factorial = 39,916,800

considering 11 places, we have that one girl is at the first place P1, then 4 boys at p2, p3, p4, p5 , then the second girl is at the 6th place and the remaining five boys take the remaining places from 7th to 11th.

P1=G, P2=B, P3= B, P4= B, P5=B, P6=G, P7=B, P8=B, P9=B, P10=B, P11=B

In this case,

2 girls can be arranged by two factorial = 2 ways

4 boys can be arranged by four factorial = 24 ways

and the remaining 5 boys can be arranged by 5 factorial = 120 ways

Hence possible number of arrangement = (2*24*120) = 5760

Now, from the above, we have that one girl is at the first place P1, then 4 boys at p2, p3, p4, p5 , then the second girl is at the 6th place and the remaining five boys take the remaining places from 7th to 11th.

In a similar way, we can place same arrangement such that the first girl is at P6, then the 4 boys at P7, P8, P9, P10, and then the second girl at 11th place P11, the remaining five boys at places P1, P2,P3,P4,P5.

Therefore, the number of ways exactly 4 boys between 2 girls = ( 6* 5760) = 34560

Hence, the required probability is defined as ( 34560/39916800) = 0.000865801 which is the required solution.