In how many ways can 10 persons take seats in a row of 24 fixed seats so that no two persons take consecutive seats.

Explanation

As no two persons take consecutive seats, there will be at least one vacant seat between any two person. Let the number of vacant seats before the first person be `x_(0)`, and the number of vacant seats between the first and the second persons be`x_(1)`, etc., as shown in Figure.

Clearly, the total number of vacant seats is 24-10=14.
`therefore x_(0)+x_(1)+x_(2)+..+ x_(9)+x_(10)=14`
where `x_(0) ge 0, x_(1) ge 1, x_(2) ge 1, x_(3) ge 1 , .., x_(9) ge 1, x_(10) ge 0`.
Let `x_(0)=y_(0),x_(1)=y_(1)+1, x_(2)=y_(2)+1, .., x_(9)=y_(9)+1, x_(10)=y_(10)`.
Then, the equation becomes
`y_(0)+(y_(1)+1)+(y_(2)+1)+..+(y_(9)+1)+y_(10)=14`
or `y_(0)+y_(1)+y_(2)+..+ y_(9)+y_(10)=14-9=5`
`therefore` Number of non-negative integral solutions of the above equation is `.^(5+11-1)C_(11-1)= .^(15)C_(10)`
But 10 persons can arrange among themselves in 10! ways. Hence, the required number of ways is
`.^(15)C_(10)xx10!=(15!)/(10!5!)xx10!=(15!)/(5!)`