Question from combinatorics. When 20 children in a classroom line up for lunch, Pat insists on being somewhere ahead of Lynn. If Pat's demand is to be satisfied, in how many ways can the children line up ?
The following is my argument.
(1) If Lynn is the last one in the row, Pat can go anywhere, so there are 19! ways to line up the children.
(2) If Lynn is second to last, then Pat has to be in front of her, so there are 18 possible children to be behind Lynn, and 18! different ways that the kids in front of her can line up.
...
after going through this argument until Pat is the 1st and Lynn is the second in the line,
the total number of ways the line can be made is
19! + 18(18!) + 18*17(17!) + ... + 18!(1)
= 18! (19 + 18 + ... 2 + 1) 
= 18!(1+19)19/2
= 20!/2
Is this an efficient way to solve this problem, or is there an easier argument that I could be making ?
It's a standardized test problem, so I cannot spend 5 min solving just one problem like this.

Pat is ahead of Lynn in exactly half of the $20!$ possible arrangements, i.e., in
$$\frac12\cdot20!=10\cdot19!$$
arrangements. Each arrangement with Pat ahead of Lynn can be paired with an arrangement that has Lynn ahead of Pat by interchanging the two and leaving everyone else alone.
