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.
 A: In any given line-up (permutation), either Pat is ahead of Lynn, or Lynn is ahead of Pat. Swapping those two children in any permutation of the first kind gives one of the second, and vice-versa. That is, there is a bijection between the two permutations. So there are an equal number of both kinds. The number of permutations of the first kind is therefore half the total number of permutations (which is $20!$), and thus $\frac{20!}{2}$.

Suppose you draw a permutation uniformly at random from all the possible permutations. By symmetry, the probability that Pat is ahead of Lynn is exactly $\frac{1}{2}$. So the number of such permutations is $\frac12(20!)$.

A more general approach, that may be useful in other problems where such symmetry arguments are not immediately apparent: Suppose you fix the positions of these two: there are $\binom{20}{2} = \frac{20\times 19}{2}$ ways of choosing these two positions. In these two positions, the places of Pat and Lynn are fixed (Pat comes first). In the other $18$ positions, the other $18$ people can be arranged in $18!$ ways, so the number of permutations is $\frac{20 \times 19}{2} 18! = \frac{20!}{2}$. Vaguely, the idea is to separate out what you're interested in (here, just these two people) from the rest.
A: 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.
A: Of course answer is 20!/2. See there are 20! ways of arranging. In half of them Pat is ahead and in the other half Pat is behind.
