Question

The number of ways in which 3 boys and 12 girls can be seated around a circle such that there are atleast 3 girls between any two is $20\times k!$, then $k$ is

Answer 1

$\text{First we seat 3 boys around the circular table.}\text{This can be done in 2! ways.}\text{Let there are}{x}_1,x_2,\text{and}{x}_3\text{girls seating between 3 boys.}\therefore x_1+x_2+x_3=12\text{such that}{x}_1\ge 3,x_2\ge 3,x_3\ge 3\text{Let}{x}_1=a+3,x_2=b+3,x_3=c+3\therefore a+b+c=3\text{such that}a\ge 0,b\ge 0,c\ge 0\text{The number of whole number solutions of this equation}{=}^5{C}_2\therefore \text{Total number of ways}{=}^5{C}_2\times 12!\times 2!=20\times 12!\therefore k=20$