If `n` objects are arrange in a row, then the number of ways of selecting three of these objects so that no two of them are next to each other is

Options

A. `""^(n-3)C_(3)`
B. `""^(n-3)C_(2)`
C. `""^(n-2)C_(3)`
D. none of these

Explanation

Let `x_(1)` be the number of objects to the left of first object, `x_(2)` the number of objects between the first and second, `x_(3)` the number of objects between second and third and `x_(4)` the number of objects to the right of the third object. Then,
`x_(1)ge0,x_(2)ge1,x_(3)ge1andx_(4)ge0` such that
`x_(1)+x_(2)+x_(3)+x_(4)=n-3" ".....(i)`
Clearly, the total number of ways is the total number of soluttions of (i).
Let `y_(2)=x_(2)-1andy_(3)=x_(3)-1` Then, equation (i) reduces to `x_(1)+y_(2)+y_(3)+x_(4)=n-5`, where `x_(1),x_(4)ge0,y_(2),y_(3)ge0`
The number of solutions of this equation is
`""^(n-5+4-1)C_(4-1)=""^(n-2)C_(3)`