In how many ways 3 boys and 15 girls can sits together in a row such that between any 2 boys at least 2 girls sit.

Answer

`3!xx15!xx.^(14)C_(3)`

Explanation

B B B
x y z w
First 3 boys can be arranged in 3! Ways. After arranging the boys, 4 gaps are created. Let in these gaps x,y,z, and w girls sit as shown in the figure. Let us first find out the distribution ways of girls in the four gaps. As given in question, `y,z ge 2 " and" x,w ge 0`, we have to find the integral solutions of the equation x+y+z+w=15 with the above condition. Let
`y=y_(1)+2 " and" z=z_(1)+2 {"where" y_(1),z_(1) ge0)`
`implies x+y_(1)+z_(1)+w=11`
Number of solutions of above equation is `.^(11+4-1)C_(4-1)= .^(14)C_(3)`. After it is decided as in which gap how many girls will sit, they can be arranged in 15! ways.
Hence, the total number of ways is `3!15! .^(14)C_(3)`